41244.A financier claims to be lending money at simple interest, But he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes.
10.25%
10%
9.25%
9%
Explanation:
Let the sum is 100.
As financier includes interest every six months., then we will calculate SI for 6 months, then again for six months as below:
SI for first Six Months = (100*10*1)/(100*2) = Rs. 5
Important: now sum will become 100+5 = 105
SI for last Six Months = (105*10*1)/(100*2) = Rs. 5.25
So amount at the end of year will be (100+5+5.25)
= 110.25
Effective rate = 110.25 - 100 = 10.25
Let the sum is 100.
As financier includes interest every six months., then we will calculate SI for 6 months, then again for six months as below:
SI for first Six Months = (100*10*1)/(100*2) = Rs. 5
Important: now sum will become 100+5 = 105
SI for last Six Months = (105*10*1)/(100*2) = Rs. 5.25
So amount at the end of year will be (100+5+5.25)
= 110.25
Effective rate = 110.25 - 100 = 10.25
41245.A sum of money amounts to Rs 9800 after 5 years and Rs 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is
9%
10%
11%
12%
Explanation:
We can get SI of 3 years = 12005 - 9800 = 2205
SI for 5 years = (2205/3)*5 = 3675 [so that we can get principal amount after deducting SI]
Principal = 12005 - 3675 = 6125
So Rate = (100*3675)/(6125*5) = 12%
We can get SI of 3 years = 12005 - 9800 = 2205
SI for 5 years = (2205/3)*5 = 3675 [so that we can get principal amount after deducting SI]
Principal = 12005 - 3675 = 6125
So Rate = (100*3675)/(6125*5) = 12%
41246.Albert invested amount of 8000 in a fixed deposit for 2 years at compound interest rate of 5 % per annum. How much Albert will get on the maturity of the fixed deposit.
Rs. 8510
Rs. 8620
Rs. 8730
Rs. 8820
Explanation:
=>(8000×(1+5100)^2)
=>8000×21/20×21/20
=>8820
=>(8000×(1+5100)^2)
=>8000×21/20×21/20
=>8820
41247.A plot is sold for Rs. 18,700 with a loss of 15%. At what price it should be sold to get profit of 15%.
Rs 25300
Rs 22300
Rs 24300
Rs 21300
Explanation:
This type of question can be easily and quickly solved as following:
Let at Rs x it can earn 15% pr0fit
85:18700 = 115:x [as, loss = 100 -15, Profit = 100 +15]
x = (18700*115)/85
= Rs.25300
This type of question can be easily and quickly solved as following:
Let at Rs x it can earn 15% pr0fit
85:18700 = 115:x [as, loss = 100 -15, Profit = 100 +15]
x = (18700*115)/85
= Rs.25300
41248.A man gains 20% by selling an article for a certain price. If he sells it at double the price, the percentage of profit will be.
130%
140%
150%
160%
Explanation:
Let the C.P. = x,
Then S.P. = (120/100)x = 6x/5
New S.P. = 2(6x/5) = 12x/5
Profit = 12x/5 - x = 7x/5
Profit% = (Profit/C.P.) * 100
=> (7x/5) * (1/x) * 100 = 140 %
Let the C.P. = x,
Then S.P. = (120/100)x = 6x/5
New S.P. = 2(6x/5) = 12x/5
Profit = 12x/5 - x = 7x/5
Profit% = (Profit/C.P.) * 100
=> (7x/5) * (1/x) * 100 = 140 %
41249.The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25% then determine the value of x.
14
15
16
17
Explanation:
Let the cost price 1 article = Re 1
Cost price of x articles = x
S.P of x articles = 20
Gain = 20 -x
=>25=(((20−x)/x)∗100)
=>2000−100x=25x
=>x=16
Let the cost price 1 article = Re 1
Cost price of x articles = x
S.P of x articles = 20
Gain = 20 -x
=>25=(((20−x)/x)∗100)
=>2000−100x=25x
=>x=16
41250.In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit
70%
80%
90%
None of above
Explanation:
Let C.P.= Rs. 100.
Then, Profit = Rs. 320,
S.P. = Rs. 420.
New C.P. = 125% of Rs. 100 = Rs. 125
New S.P. = Rs. 420.
Profit = Rs. (420 - 125) = Rs. 295
Required percentage = (295/420) * 100
= 70%(approx)
Let C.P.= Rs. 100.
Then, Profit = Rs. 320,
S.P. = Rs. 420.
New C.P. = 125% of Rs. 100 = Rs. 125
New S.P. = Rs. 420.
Profit = Rs. (420 - 125) = Rs. 295
Required percentage = (295/420) * 100
= 70%(approx)
41251.A man bought an article and sold it at a gain of 5 %. If he had bought it at 5% less and sold it for Re 1 less, he would have made a profit of 10%. The C.P. of the article was
Rs 100
Rs 150
Rs 200
Rs 250
Explanation:
Let original Cost price is x
Its Selling price = 105/100 * x = 21x/20
New Cost price = 95/100 * x = 19x/20
New Selling price = 110/100 * 19x/20 = 209x/200
[(21x/20) - (209x/200)] = 1
=> x = 200
Let original Cost price is x
Its Selling price = 105/100 * x = 21x/20
New Cost price = 95/100 * x = 19x/20
New Selling price = 110/100 * 19x/20 = 209x/200
[(21x/20) - (209x/200)] = 1
=> x = 200
41252.A fruit seller sells mangoes at the rate of Rs.9 per kg and thereby loses 20%. At what price per kg, he should have sold them to make a profit of 5%
Rs 8.81
Rs 9.81
Rs 10.81
Rs 11.81
Explanation:
85 : 9 = 105 : x
x= (9×105/85)
= Rs 11.81
85 : 9 = 105 : x
x= (9×105/85)
= Rs 11.81
41253.A shopkeeper sold an article for Rs 2564.36. Approximately what was his profit percent if the cost price of the article was Rs 2400
4%
5%
6%
7%
Explanation:
Gain % = (164.36*100/2400) = 6.84 % = 7% approx
Gain % = (164.36*100/2400) = 6.84 % = 7% approx
41254.A and B can together finish a work 30 days. They worked together for 20 days and then B left. After another 20 days, A finished the remaining work. In how many days A alone can finish the work?
40
50
54
60
Explanation:
(A + B)s 20 days work = (1/30 x 20 ) = 2/3 .
Remaining work = (1 - 2/3) = 1/3 .
Now, 1/3 work is done by A in 20 days.
Therefore, the whole work will be done by A in (20 x 3) = 60 days.
(A + B)s 20 days work = (1/30 x 20 ) = 2/3 .
Remaining work = (1 - 2/3) = 1/3 .
Now, 1/3 work is done by A in 20 days.
Therefore, the whole work will be done by A in (20 x 3) = 60 days.
41255.A person incurs a loss of 5% be selling a watch for Rs. 1140. At what price should the watch be sold to earn 5% profit.
Rs.1200
Rs.1230
Rs.1260
Rs.1290
Explanation:
Let the new S.P. be x, then.
(100 - loss%):(1st S.P.) = (100 + gain%):(2nd S.P.)
=>(95/1140=105/x)=>x=1260
Let the new S.P. be x, then.
(100 - loss%):(1st S.P.) = (100 + gain%):(2nd S.P.)
=>(95/1140=105/x)=>x=1260
41256.A book was sold for Rs 27.50 with a profit of 10%. If it were sold for Rs. 25.75, then would have been percentage of profit and loss ?
2% Profit
3% Profit
2% Loss
3% Loss
Explanation:
Please remember
S.P.=(((100+gain%)/100)∗C.P)
So, C.P. = ((100/110)∗25.75)
When S.P. = 25.75 then Profit=25.75−25=Re.0.75
Profit%=(0.75/25)∗100=3%
Please remember
S.P.=(((100+gain%)/100)∗C.P)
So, C.P. = ((100/110)∗25.75)
When S.P. = 25.75 then Profit=25.75−25=Re.0.75
Profit%=(0.75/25)∗100=3%
41257.If the cost price is 25% of selling price. Then what is the profit percent.
150%
200%
300%
350%
Explanation:
Let the S.P = 100
then C.P. = 25
Profit = 75
Profit% = 75/25 * 100 = 300%
Let the S.P = 100
then C.P. = 25
Profit = 75
Profit% = 75/25 * 100 = 300%
41258.The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, find out the value of x
13
14
15
16
Explanation:
Let the Cost Price of one article = Rs. 1
CP of x articles = Rs. x
CP of 20 articles = 20
Selling price of x articles = 20
Profit = 25% [Given]
⇒((SP−CP)/CP) = 25/100=1/4⇒(20−x)/x=¼
⇒80−4x=x
⇒5x=80
⇒x=80/5=16
Let the Cost Price of one article = Rs. 1
CP of x articles = Rs. x
CP of 20 articles = 20
Selling price of x articles = 20
Profit = 25% [Given]
⇒((SP−CP)/CP) = 25/100=1/4⇒(20−x)/x=¼
⇒80−4x=x
⇒5x=80
⇒x=80/5=16
41259.A man buys an item at Rs. 1200 and sells it at the loss of 20 percent. Then what is the selling price of that item
Rs. 660
Rs. 760
Rs. 860
Rs. 960
Explanation:
Here always remember, when ever x% loss,
it means S.P. = (100 - x)% of C.P
when ever x% profit,
it means S.P. = (100 + x)% of C.P
So here will be (100 - x)% of C.P.
= 80% of 1200
= 80/100 * 1200
= 960
Here always remember, when ever x% loss,
it means S.P. = (100 - x)% of C.P
when ever x% profit,
it means S.P. = (100 + x)% of C.P
So here will be (100 - x)% of C.P.
= 80% of 1200
= 80/100 * 1200
= 960
41260.Sahil purchased a machine at Rs 10000, then got it repaired at Rs 5000, then gave its transportation charges Rs 1000. Then he sold it with 50% of profit. At what price he actually sold it.
Rs. 22000
Rs. 24000
Rs. 26000
Rs. 28000
Explanation:
Question seems a bit tricky, but it is very simple.
Just calculate all Cost price, then get 150% of CP.
C.P. = 10000 + 5000 + 1000 = 16000
150% of 16000 = 150/100 * 16000 = 24000
Question seems a bit tricky, but it is very simple.
Just calculate all Cost price, then get 150% of CP.
C.P. = 10000 + 5000 + 1000 = 16000
150% of 16000 = 150/100 * 16000 = 24000
41261.A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?
35 ½
36 ½
37 ½
38 ½
Explanation:
Work done by A in 20 days = 80/100 = 8/10 = 4/5
Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 --- (1)
Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)
Work done by A and B in 1 day = 1/15 ---(2)
Work done by B in 1 day = 1/15 – 1/25 = 2/75
=> B can complete the work in 75/2 days = 37 (1/2) days
Work done by A in 20 days = 80/100 = 8/10 = 4/5
Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 --- (1)
Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)
Work done by A and B in 1 day = 1/15 ---(2)
Work done by B in 1 day = 1/15 – 1/25 = 2/75
=> B can complete the work in 75/2 days = 37 (1/2) days
41262.4 men and 6 women finish a job in 8 days, while 3 men and 7 women finish it in 10 days. In how many days will 10 women working together finish it ?
30 days
40 days
50 days
60 days
Explanation:
Let 1 mans 1 day work = x
and 1 womans 1 days work = y.
Then, 4x + 6y = 1/8
and 3x+7y = 1/10
solving, we get y = 1/400 [means work done by a woman in 1 day]
10 women 1 day work = 10/400 = 1/40
10 women will finish the work in 40 days
Let 1 mans 1 day work = x
and 1 womans 1 days work = y.
Then, 4x + 6y = 1/8
and 3x+7y = 1/10
solving, we get y = 1/400 [means work done by a woman in 1 day]
10 women 1 day work = 10/400 = 1/40
10 women will finish the work in 40 days
41263.10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?
6 days
7 days
8 days
9 days
Explanation:
1 womans 1 days work = 1/70
1 Childs 1 days work = 1/140
5 Women and 10 children 1 day work =
(5/70 +10/140)=1/7
So 5 women and 10 children will finish the work in 7 days.
1 womans 1 days work = 1/70
1 Childs 1 days work = 1/140
5 Women and 10 children 1 day work =
(5/70 +10/140)=1/7
So 5 women and 10 children will finish the work in 7 days.
41264.5 men and 2 boys working together can do four times as much work as a man and a boy. Working capacity of man and boy is in the ratio
1:2
1:3
2:1
2:3
Explanation:
Let 1 man 1 day work = x
1 boy 1 day work = y
then 5x + 2y = 4(x+y)
=> x = 2y
=> x/y = 2/1
=> x:y = 2:1
Let 1 man 1 day work = x
1 boy 1 day work = y
then 5x + 2y = 4(x+y)
=> x = 2y
=> x/y = 2/1
=> x:y = 2:1
41265.Rahul and Sham together can complete a task in 35 days, but Rahul alone can complete same work in 60 days. Calculate in how many days Sham can complete this work ?
84 days
82 days
76 days
68 days
Explanation:
As Rahul and Sham together can finish work in 35 days.
1 days work of Rahul and Sham is 1/35
Rahul can alone complete this work in 60 days,
So, Rahul one day work is 1/60
Clearly, Sham one day work will be = (Rahul and Sham one day work) - (Rahul one day work)
=1/35−1/60=1/84
Hence Sham will complete the given work in 84 days.
As Rahul and Sham together can finish work in 35 days.
1 days work of Rahul and Sham is 1/35
Rahul can alone complete this work in 60 days,
So, Rahul one day work is 1/60
Clearly, Sham one day work will be = (Rahul and Sham one day work) - (Rahul one day work)
=1/35−1/60=1/84
Hence Sham will complete the given work in 84 days.
41266.If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
4 days
5 days
6 days
7 days
Explanation:
Let 1 mans 1 days work = x and 1 boys 1 days work = y.
Then, 6x + 8y = 1/10 and 26x + 48y = 1/2 .
Solving these two equations, we get : x = 1/100 and y = 1/200 .
(15 men + 20 boy)s 1 days work = 15/100 + 20/200 = 1/4 .
15 men and 20 boys can do the work in 4 days.
Let 1 mans 1 days work = x and 1 boys 1 days work = y.
Then, 6x + 8y = 1/10 and 26x + 48y = 1/2 .
Solving these two equations, we get : x = 1/100 and y = 1/200 .
(15 men + 20 boy)s 1 days work = 15/100 + 20/200 = 1/4 .
15 men and 20 boys can do the work in 4 days.
41267.There are three numbers, these are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. What will be the sum of three numbers :
80
82
85
87
Explanation:
As given the questions these numbers are co primes, so there is only 1 as their common factor.
It is also given that two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
So first number is : 551/29 = 19
Third number = 1073/29 = 37
So sum of these numbers is = (19 + 29 + 37) = 85
As given the questions these numbers are co primes, so there is only 1 as their common factor.
It is also given that two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
So first number is : 551/29 = 19
Third number = 1073/29 = 37
So sum of these numbers is = (19 + 29 + 37) = 85
41268.The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
123
127
235
305
Explanation:
Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.
Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.
41269.If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:
55/601
601/55
11/120
120/11
Explanation:
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum = 1/a + 1/b = a + b/ab = 55/600 = 11/120
Let the numbers be a and b.
Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum = 1/a + 1/b = a + b/ab = 55/600 = 11/120
41270.Choose the correct answer.
What is the value of $4^-2 $?
What is the value of $4^-2 $?
Option 1 : 1/4
Option 2 : 1/16
Option 3 : -16
Option 4 : None of these
41271.Ques : Choose the correct answer.
What is the value of (0.081)1/4?
What is the value of (0.081)1/4?
Option 1 : 0.3
Option 2 : 0.03
Option 3 : 0.003
Option 4 : None of these
41272.A tyre has two punctures. The first puncture alone would have made the tyre flat in 9 minutes and the second alone would have done it in 6 minutes. If air leaks out at a constant rate, how long does it take both the punctures together to make it flat ?
3 ⅕ min
3 ⅖ min
3 ⅗ min
3 ⅘ min
Explanation:
Do not be confused, Take this question same as that of work done questions. Like work done by 1st puncture in 1 minute and by second in 1 minute.
Lets Solve it:
1 minute work done by both the punctures =
(1/9+1/6)=(5/18)
So both punctures will make the type flat in
(18/5)mins=3 ⅗ mins
Do not be confused, Take this question same as that of work done questions. Like work done by 1st puncture in 1 minute and by second in 1 minute.
Lets Solve it:
1 minute work done by both the punctures =
(1/9+1/6)=(5/18)
So both punctures will make the type flat in
(18/5)mins=3 ⅗ mins
41273.A is twice as good as workman as B and together they finish a piece of work in 18 days. In how many days will B alone finish the work.
27 days
54 days
56 days
68 days
Explanation:
As per question, A do twice the work as done by B.
So A:B = 2:1
Also (A+B) one day work = 1/18
To get days in which B will finish the work, lets calculate work done by B in 1 day =
=(1/18∗1/3)=1/54
[Please note we multiplied by 1/3 as per B share and total of ratio is 1/3]
So B will finish the work in 54 days
As per question, A do twice the work as done by B.
So A:B = 2:1
Also (A+B) one day work = 1/18
To get days in which B will finish the work, lets calculate work done by B in 1 day =
=(1/18∗1/3)=1/54
[Please note we multiplied by 1/3 as per B share and total of ratio is 1/3]
So B will finish the work in 54 days
41274.To complete a work A and B takes 8 days, B and C takes 12 days, A,B and C takes 6 days. How much time A and C will take
24 days
16 days
12 days
8 days
Explanation:
A+B 1 day work = 1/8
B+C 1 day work = 1/12
A+B+C 1 day work = 1/6
We can get A work by (A+B+C)-(B+C)
And C by (A+B+C)-(A+B)
So A 1 day work =
1/6−1/12=1/12
Similarly C 1 day work =
1/6−1/8=(4−3)/24=1/24
So A and C 1 day work =
1/12+1/24=3/24=1/8
So A and C can together do this work in 8 days
A+B 1 day work = 1/8
B+C 1 day work = 1/12
A+B+C 1 day work = 1/6
We can get A work by (A+B+C)-(B+C)
And C by (A+B+C)-(A+B)
So A 1 day work =
1/6−1/12=1/12
Similarly C 1 day work =
1/6−1/8=(4−3)/24=1/24
So A and C 1 day work =
1/12+1/24=3/24=1/8
So A and C can together do this work in 8 days
41275.A can do a piece of work in 15 days and B alone can do it in 10 days. B works at it for 5 days and then leaves. A alone can finish the remaining work in
5 days
6 days
7.5 days
8.5 days
Explanation:
Bs 5 days work =
1/10∗5=½
Remaining work =1−1/2=½
A can finish work =15∗1/2=7.5days
Bs 5 days work =
1/10∗5=½
Remaining work =1−1/2=½
A can finish work =15∗1/2=7.5days
41276.Find the largest number which divides 62,132,237 to leave the same reminder
30
32
35
45
Explanation:
Trick is HCF of (237-132), (132-62), (237-62)
= HCF of (70,105,175) = 35
Trick is HCF of (237-132), (132-62), (237-62)
= HCF of (70,105,175) = 35
41277.Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
4
5
6
8
Explanation:
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
41278.The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
101
107
111
185
Explanation:
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
41279.Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
40
80
120
200
Explanation:
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
41280.The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
3
13
23
33
Explanation:
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 - 37) = 23.
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 - 37) = 23.
41281.A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?
26 minutes and 18 seconds
42 minutes and 36 seconds
45 minutes
46 minutes and 12 seconds
Explanation:
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
41282.Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds resp. In 60 minutes how many times they will toll together.
15
16
30
31
Explanation:
LCM of 2-4-6-8-10-12 is 120 seconds, that is 2 minutes.
Now 60/2 = 30
Adding one bell at the starting it will 30+1 = 31
LCM of 2-4-6-8-10-12 is 120 seconds, that is 2 minutes.
Now 60/2 = 30
Adding one bell at the starting it will 30+1 = 31
41283.If the sum of two numbers is 55 and the H.C.F. and L.C.M of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:
55/601
601/55
11/120
120/11
41284.Three different containers contain 496 litres, 403 litres and 713 litres of mixtures of milk and water respectively. What biggest measure can measure all the different quantities exactly ?
1 litre
7 litre
31 litre
41 litre
41285.Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?
4
10
15
16
41286.Four different electronic devices make a beep after every 30 minutes, 1 hour, 3/2 hour and 1 hour 45 minutes respectively. All the devices beeped together at 12 noon. They will again beep together at:
12 midnight
3 a.m.
6 a.m.
9 a.m.
41290.The number nearest to 15207, which is divisible by 467, is:
14342
15211
14944
15411
None of these
41291.The smallest number, which is a perfect square and contains 7936 as a factor is:
251664
231564
246016
346016
None of these
41292.In a division problem, the divisor is twenty times the quotient and five times the remainder. If remainder is 16, the number will be:
3360
336
1616
20516
None of these
41293.The L.C.M. of two numbers is 4800 and their G.C.M. is 160. If one of the numbers is 480, then the other number is:
1600
1800
2200
2600
None of these