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Aptitude Simplification Practice QA

3184.To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two-fifth of its present ?
10
35
62.5
Cannot be determined
Explanation:

Let the capacity of 1 bucket = $ x $.

Then, the capacity of tank = 25$ x $.

New capacity of bucket =$ \dfrac{2}{5} x $

$\therefore$ Required number of buckets =$ \dfrac{25x}{(2x/5)} $

=$ \left( 25x\times\dfrac{5}{2x} \right) $

=$ \dfrac{125}{2} $

= 62.5

3186.The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:
Rs. 3500
Rs. 3750
Rs. 3840
Rs. 3900
Explanation:

Let the cost of a chair and that of a table be Rs. $ x $ and Rs. $ y $ respectively.

Then, 10$ x $ = 4$ y $   or   $ y $ =$ \dfrac{5}{2} x $.

$\therefore$ 15$ x $ + 2$ y $ = 4000

$\Rightarrow$ 15$ x $ + 2$ \times \dfrac{5}{2} x $ = 4000

$\Rightarrow$ 20$ x $ = 4000

$\therefore x $ = 200.

So, $ y $ =$ \left(\dfrac{5}{2} \times 200\right) $= 500.

Hence, the cost of 12 chairs and 3 tables = 12$ x $ + 3$ y $

    = Rs. $\left(2400 + 1500\right)$

    = Rs. 3900.

3188.David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross ?
19
28
30
37
Explanation:

Suppose their paths cross after $ x $ minutes.

Then, 11 + 57$ x $ = 51 - 63$ x $    $\Leftrightarrow$    120$ x $ = 40

x = $ \dfrac{1}{3} $

Number of floors covered by David in $\left(1/3\right)$ min. =$ \left(\dfrac{1}{3} \times 57\right) $= 19.

So, their paths cross at $\left(11 +19\right)$i.e., 30th floor.

3189.One-third of Rahuls savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,50,000 as total savings, how much has he saved in Public Provident Fund ?
Rs. 30,000
Rs. 50,000
Rs. 60,000
Rs. 90,000
Explanation:

Let savings in N.S.C and P.P.F. be Rs. $ x $ and Rs. $\left(150000 - x \right)$ respectively. Then,

$ \dfrac{1}{3} $x =$ \dfrac{1}{2} \left(150000 - x \right)$

$\Rightarrow \dfrac{x}{3} $+$ \dfrac{x}{2} $= 75000

$\Rightarrow \dfrac{5x}{6} $= 75000

$\Rightarrow x $ =$ \dfrac{75000 \times 6}{5} $= 90000

$\therefore$ Savings in Public Provident Fund = Rs. $\left(150000 - 90000\right)$ = Rs. 60000

3191.In a group of ducks and cows, the total number of legs are 28 more than twice the number of heads. Find the total number of cows.
14
12
16
8
Explanation:

Let the number of ducks be d

and number of cows be c

Then, total number of legs = 2d + 4c = 2$\left(d + 2c\right)$

total number of heads = c + d

Given that total number of legs are 28 more than twice the number of heads

=> 2(d + 2c) = 28 + 2(c + d)

=> d + 2c = 14 + c + d

=> 2c = 14 + c

=> c = 14

i.e., total number of cows = 14

3195.A fires 5 shots to Bs 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:
30 birds
60 birds
72 birds
90 birds
Explanation:

Let the total number of shots be $ x $. Then,

Shots fired by A =$ \dfrac{5}{8} x $

Shots fired by B =$ \dfrac{3}{8} x $

Killing shots by A =$ \dfrac{1}{3} $of$ \dfrac{5}{8} x $=$ \dfrac{5}{24} x $

Shots missed by B =$ \dfrac{1}{2} $of$ \dfrac{3}{8} x $=$ \dfrac{3}{16} x $

$\therefore \dfrac{3x}{16} $= 27 or $ x $ =$ \left(\dfrac{27 \times 16}{3} \right) $= 144.

Birds killed by A =$ \dfrac{5x}{24} $=$ \left(\dfrac{5}{24} \times 144\right) $= 30.

10711.240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (11 × 24 ÷ 8 × 3 – 69) × 2}
3841
3855
3441
3900
Explanation:

240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (11 × 24 ÷ 8 × 3 – 69) × 2}
= 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (264 ÷ 8 × 3 – 69) × 2}
= 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (33 × 3 – 69) × 2}
= 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ (99 – 69) × 2}
= 240 ÷ 8 × 512 ÷ 4 + ½ of {1800 ÷ 30 x 2}
= 240 ÷ 8 × 512 ÷ 4 + ½ of {60 x 2}
= 240 ÷ 8 × 512 ÷ 4 + ½ of 120
= 240 ÷ 8 × 512 ÷ 4 + ½ × 120
= 240 ÷ 8 × 512 ÷ 4 + 60
= 30 × 512 ÷ 4 + 60
= 15360 ÷ 4 + 60
= 3840 + 60
= 3900


10714.25 – 1/2{5 + 4 – (3 + 2 – 1 + 3)}
24
21
6
16
Explanation:

25 – 1/2{5 + 4 – (3 + 2 – 1 + 3)}

= 25 – 1/2{5 + 4 – 7}

= 25 – 1/2*2

= 25 –1

= 24

10715.27 – [38 – {46 – (15 – 13 – 2)}]
19
25
31
35
Explanation:

27 – [38 – {46 – (15 – 13 – 2)}]

= 27 – [38 – {46 – 0}]

=27 – [38 – 46]

=27 – [–8]

=27 + 8


=35

10719.2550 – [510 – {270 – (90 – 80 + 70)}]
2270
2370
2230
2330
Explanation:

2550 – [510 – {270 – (90 – 80 + 70)}]

= 2550 – [510 – {270 – (10 + 70)}]

= 2550 – [510 – {270 – 80}]

= 2550 – [510 – 190]

= 2550 – 320

= 2230

10720.4 + (1/5) [{–10 x (25 – 13 – 3)} ÷ (–5)]
7.6
10
8.2
18
Explanation:

4 + (1/5) [{–10 x (25 – 13 – 3)} ÷ (–5)]

= 4 + (1/5) [{–10 x 9} ÷ (–5)]

= 4 + (1/5) [–90 ÷ (–5)]

= 4 + (1/5) [18]

= 4 + 1 ÷ 5 × 18


= 4 + .2 × 18

= 4 + 3.6


= 7.6

10721.22 – (1/4) {–5 – (– 48) ÷ (–16)}
19
20
24
21
Explanation:

22 – (1/4) {–5 – (– 48) ÷ (–16)}

= 22 – (1/4) {–5 – 3}

= 22 – (1/4) × –8

= 22 – 1 × –2

= 22 + 2

= 24

10722.63 – (–3) {–2 – 8 – 3} ÷ 3{5 + (–2) (–1)}
–1820
62
–28
63
Explanation:

63 – (–3) {–2 – 8 – 3} ÷ 3{5 + (–2) (–1)}

= 63 – (–3) {–2 – 8 – 3} ÷ 3{5 + 2}

= 63 – (–3) {–2 – 8 – 3} ÷ 3 × 7

= 63 – (–3) × (–13) ÷ 3 × 7


= 63 – 39 ÷ 3 × 7

= 63 – 13 × 7

= 63 – 91

= –28

10723.[29 – (–2) {6 – (7 – 3)}] ÷ [3 x {5 + (–3) x (–2)}]
1
–1
0
2
Explanation:

[29 – (–2) {6 – (7 – 3)}] ÷ [3 x {5 + (–3) x (–2)}]

= [29 – (–2) {6 – 4}] ÷ [3 x {5 + 6}]

= [29 – (–2) × 2] ÷ [3 x 11]

= [29 – (–4)] ÷ 33

= [29 + 4] ÷ 33

= 33 ÷ 33

= 1


10726.Find the value of 1/(3+1/(3+1/(3–1/3)))
3/10
10/3
27/89
89/27
Explanation:

1/(3+1/(3+1/(3–1/3)))

= 1/(3 + 1/(3 + 1/(8/3)))

= 1/(3 + 1/(3 + 3/8))

= 1/(3 + 8/27)

= 1/(89/27)

= 27/89

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