6x - 27 + 3x = 4 + 9 - x
or, 10x = 40
or, x = 4
Given:-
The cost of 2 chairs and 3 tables = Rs.1300
The cost of 3 chairs and 2 tables = Rs.1200
To Find:-
By how much is the cost of table more than the cost of chair.
Solution:-
Let the cost of each chair be x
And cost of each table be y
Case 1:-
Cost of 2 chairs + 3 tables = 1300
Case 2:-
Cost of 3 chairs + 2 tables = 1200
Multiply equation (i) with 3 and (ii) with 2
Equation (iii) - (iv)
Substitute y = 300 in equation (i)
Therefore:-
Cost of a chair = Rs.200
Cost of a table = Rs.300
Hence,
The cost of each table is more than that of each chair by Rs.100
Given,
Cost of 10kg apples=cost of 24kg rice
Cost of 6kg flour=cost of 2kg rice
Cost of 1kg apples=20.50
Cost of 4kg apples +3kg rice + 5kg flour=?
Solution:
6 kg of flour= 1 kg flour $\times 6$
=20.50 $\times 6$
=Rs.123.
Cost of 6kg flour=cost of 2kg rice
Therefore,2 kg rice =Rs.123.
Then the 24kg of rice =$12 \times 123$ =1476
Cost of 10kg apples=cost of 24kg rice
So,10 kg of apples=Rs.1476
Then the 4kg apples =$\dfrac{1476}{10}$=$147.6 \times 4 $=Rs.590.4
Then, 3 kg of rice=123/2=$61.5 \times 3$=Rs.184.5
Then,5 kg of flour =$20.50 \times 5$=Rs.102.50
Therefore,Total Cost of 4kg apples +3kg rice + 5kg flour
=Rs.590.4+Rs.184.5 +Rs.102.50=Rs.877.40
It is not possible to compare A and B as the data provided is inadequate.
Let the number of hens be $ x $ and the number of cows be $ y $.
Then, $ x $ + $ y $ = 48 .... (i)
and 2$ x $ + 4$ y $ = 140 $\Rightarrow$ $ x $ + 2$ y $ = 70 .... (ii)
Solving (i) and (ii) we get: $ x $ = 26, $ y $ = 22.
$\therefore$ The required answer = 26.
b - [b -(a+b) - {b - (b - a+b)} + 2a]
= b - [b -a -b - {b -b +a -b} + 2a]
= b - [b -a -b - {a -b} + 2a]
= b - [b -a -b - a +b + 2a]
= b - [b]
= b - b
= 0
Let number of notes of each denomination be $ x $.
Then $ x $ + 5$ x $ + 10$ x $ = 480
$\Rightarrow$ 16$ x $ = 480
$\therefore x $ = 30.
Hence, total number of notes = 3$ x $ = 90.
a * b = 2a - 4b + 2ab
Hence,
2*3 = 2$(2)$ - 4$(3)$ + 2$\left(2 \times 3\right)$ = 4 - 12 + 12 = 4
3*2 = 2$(3)$ - 4$(2)$ + 2$\left(3 \times 2\right)$ = 6 - 8 + 12 = 10
∵ 2*3 + 3*2 = 4 + 10 = 14
Original share of 1 person =$ \dfrac{1}{8} $
New share of 1 person =$ \dfrac{1}{7} $
Increase =$ \left(\dfrac{1}{7} -\dfrac{1}{8} \right) $=$ \dfrac{1}{56} $
$\therefore$ Required fraction =$ \dfrac{(1/56)}{(1/8)} $=$ \left(\dfrac{1}{56} \times\dfrac{8}{1} \right) $=$ \dfrac{1}{7} $
$\text{Required number of employees = }\dfrac{\text{(624000 - 600000)}}{60} = \dfrac{24000}{60} = 400$
Let the number of notes of each denomination be x
Then, x + 5x + 20x = 312
=> 26x = 312
=> x = 312/26 = 12
the total number of notes that he has = 3x = 3 × 12 = 36
Between the 12 mango trees, there are 11 gaps and each gap has 2 meter length
Also, 1 meter is left from all sides of the boundary of the garden.
Hence, length of the garden = $\left(11 \times 2\right)$ + 1 + 1 = 24 meter
Let the number of girls = x and the number of boys = 8x
Then, total number of students = x + 8x = 9x
i.e., the total number of students must be a multiple of 9
From the given choices, 42 cannot be a multiple of 9.
Hence, 42 cannot be the total number of students.
Let total number of children be $ x $.
Then, $ x \times \dfrac{1}{8} x $ = $\left(\dfrac{x}{2}\times 16\right)$ $\Leftrightarrow $ x = 64.
$\therefore$ Number of notebooks =$ \dfrac{1}{8} x $2 =$ \left(\dfrac{1}{8} \times 64 \times 64\right)$= 512.
Assume that initial number of men = initial number of women = x
2$\left(x-8\right)$ = x
=> 2x - 16 = x
=> x = 16
Total number of men and women = 2x = $2 \times 16$ = 32
– 25 + 14 ÷ (5 – 3)
= –25 + 14 ÷ 2
= – 25 + 7
= –18
36 – [18 – {14 – (15 – 4 ÷ 2 x 2)}]
= 36 – [18 – {14 – (15 – 2 x 2)}]
= 36 – [18 – {14 – (15 – 4)}]
= 36 – [18 – {14 – 11}]
= 36 – [18 – 3]
= 36 – 15
= 21
45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
= 45 – [38 – {60 ÷ 3 – (6 – 3) ÷ 3}]
= 45 – [38 – {60 ÷ 3 – 3 ÷ 3}]
= 45 – [38 – {20 – 3 ÷ 3}]
= 45 – [38 – {20 – 1}]
= 45 – [38 – 19]
= 45 – 19
= 26