Practice Finding sum difference product problems - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) What number should be subtracted from both the terms of the ratio 11 : 15 so as to make it as 2 : 3 ?
(a)
(b)
(c)
(d)
Required number = x
${11 - x}/{15 - x} = 2/3$
33 - 3x = 30 - 2x
3x - 2x = 33 - 30
x = 3
Q-2) The ratio between two numbers is 2 : 3. If each number is increased by 4, the ratio between them becomes 5 : 7. The difference between the numbers is
(a)
(b)
(c)
(d)
Let the numbers be 2x and 3x.
${2x + 4}/{3x + 4} = 5/7$
15x + 20 = 14x + 28
x = 28 - 20 = 8
Required difference
Using Rule 34,
Here, a = 2, b = 3,c = 5
d = 7 and x = 4
1st Number = ${xa(c-d)}/{ad-bc}$
= ${4 ×2(5 - 7)}/{2 × 7 - 5 × 3}$
= ${8 × - 2}/{14 - 15}$ = 16
2nd Number= ${xb(c-d)}/{ad-bc}$
= ${4 ×3(5 - 7)}/{2 × 7 - 5 × 3}$
= ${4 × 3 (- 2)}/{14 - 15}$ = 24
Difference of numbers = 24 - 16 = 8
Q-3) Three numbers are in the ratio 1 : 2 : 3 and the sum of their cubes is 4500 . The smallest number is
(a)
(b)
(c)
(d)
Let the numbers be x, 2x and 3x.
According to the question,
$x^3 + (2x)^3 + (3x)^3$ = 4500
$x^3 + 8x^3 + 27x^3$ = 4500
$36x^3$ = 4500
$x^3 = 4500/36$ = 125
$x = √^3{125}$= 5 = smallest number
Q-4) Two numbers are in the ratio of 3 : 5. If 9 is subtracted from each then they are in the ratio 12 : 23. The smaller number is
(a)
(b)
(c)
(d)
Numbers = 3x and 5x (let)
According to question,
${3x - 9}/{5x - 9} = 12/23$
69x - 207 = 60x - 108
69x - 60x = 207 - 108
9x = 99 ⇒ $x = 99/9$ = 11
Smaller number = 3x = 3 × 11 = 33
Q-5) The ratio of number of boys to the number of girls in a school of 432 pupils is 5 : 4. When some new boys and girls are admitted, the number of boys increase by 12 and the ratio of the boys to girls changes to 7 : 6. The number of new girls admitted is
(a)
(b)
(c)
(d)
Original number of boys in school
= $5/9$ × 432 = 240
Number of girls
= 432 - 240 = 192
Let the new number of girls be x.
According to the question,
${240 + 12}/{192 + x} = 7/6$
$252/{192 + x} = 7/6$
192 × 7 + 7x = 252 × 6
1344 + 7x = 1512
7x = 1512 - 1344 = 168
$x = 168/7$ = 24
Q-6) Two numbers are in the ratio 3 : 5. If each number is increased by 10, the ratio becomes 5 :7. The smaller number is
(a)
(b)
(c)
(d)
Let the numbers be 3x and 5x.
${3x + 10}/{5x + 10} = 5/7$
25x + 50 = 21x + 70
4x = 20 ⇒ x = 5
Smaller number
= 3x = 3 × 5 = 15
Using Rule 34,
Here, a = 3, b = 5, c = 5, d = 7, x = 10
Smallest number = ${xa(c-d)}/{ad-bc}$ a < b
= ${10 × 3(5 - 7)}/{3 × 7 - 5 × 5}$
= ${- 60}/{21 - 25} = 60/4$ = 15
Q-7) The number of students in three classes are in the ratio 2 : 3 : 4. If 12 students are increased in each class, this ratio changes to 8 : 11 :14. The total number of students in the three classes at the beginning was
(a)
(b)
(c)
(d)
Let the original number of students be 2x, 3x and 4x in three class.
According to the question,
${2x + 12}/{3x + 12} = 8/11$
24x + 96 = 22x + 132
2x = 132 - 96 = 36
$x = 36/2$ = 18
Original number of students
= 2x + 3x + 4x
= 9x = 9 × 18 = 162
Q-8) The ratio of two positive numbers is 3 : 4. The sum of their squares is 400. What is the sum of the numbers ?
(a)
(b)
(c)
(d)
Let two positive numbers be 3x and 4x.
According to the question,
$(3x)^2 + (4x)^2$ = 400
$9x^2 + 16x^2$ = 400
$25x^2$ = 400
$x^2 = 400/25$ = 16
$x = √16$ = 4
Sum of numbers
= 3x + 4x = 7x
= 7 × 4 = 28
Q-9) Ram got twice as many marks in English as in Science. His total marks in English, Science and Maths are 180. If the ratio of his marks in English and Maths is 2 : 3, what is his marks in Science ?
(a)
(b)
(c)
(d)
Marks in English = 2x
Marks in Maths = 3x
Marks in Science = x
x + 2x + 3x = 180
6x = 180 ⇒ x = 30
Q-10) The ratio of number of boys to that of girls in a group becomes 2:1 when 15 girls leave. But, afterwards, when 45 boys also leave, the ratio becomes 1 : 5. Originally the number of girls in the group was
(a)
(b)
(c)
(d)
Let the original number of boys and girls be x and y respectively.
Then $x/{y - 15} = 2/1$
x = 2y - 30 ....(i)
Again, ${x - 45}/{y - 15} =1/5$
5x - 225 = y - 15
5x = y - 15 + 225
5 (2y - 30) = y + 210
[From equation (i)]
10y - 150 = y + 210
10y - y = 210 + 150
9y = 360 ⇒ $y = 360/9$ = 40