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# Maths Questions for CTET Exam 2017

Q1. The ratio of the present ages of Pawan and Riya are 9 : 6. After 7 years their ages will be in the ratio of 4 : 3. Find the ratio of the sum and difference of the present ages of Pawan and Riya.
(a) 15 : 3
(b) 20 : 17
(c) 3 : 5
(d) 16 : 5
Q2. Aman sister was three times as old as Aman, ten years ago. After ten years she will be twice as old as Aman. Then Aman present age is.
(a) 20years
(b) 30years
(c) 40years
(d) 35years
Q3. Ajay mother’s age is three times as much as the sum of the ages of his three children but 5 years hence his age will be only double the sum of their ages. Then the age of the mother is:
(a) 65 years
(b) 39years
(c) 50years
(d) 75years
Q4. Anita age is 150% of what it was 10 years ago, but 75% of what it will be after 10 years. What is her present age?
(a) 20 years
(b) 30 years
(c) 40 years
(d) 45 years
Q5. If each side of a rectangle is increased by 25%, then its area will increase by
(a) 10.5%
(b) 12.5%
(c) 56.25%
(d) 25.5%
Q6. The ratio of the cost price and selling price of a toy is 25 : 24. The loss per cent is
(a) 2
(b) 5
(c) 4
(d) 7
Q7. A cloth is sold at 4% loss. Then, the ratio of selling price and cost price will be
(a) 26 : 25
(b) 25 : 26
(c) 23 : 24
(d) 24 : 25
Q8. By selling a book for Rs. 800 a man lost 20%. At what price should he have sold it to gain 20%?
(a) Rs. 1200
(b) Rs. 1350
(c) Rs. 1300
(d) Rs. 1600
Q9. If an earphone is sold for Rs. 188 at a loss of 6%, what should be its selling price in order to earn a profit of 6%?
(a) Rs. 212
(b) Rs. 250
(c) Rs. 112
(d) Rs. 121
Q10. Neha sells some oranges to Sneha making a profit of 1/6 of his outlay. Sneha sells it to Tanya, gaining 10%. If Tanya sells if for Rs. 700 and incurs a loss of 1/4 of his outlay, the cost price of Neha is approx.
(a) Rs. 628
(b) Rs. 727
(c) Rs. 772
(d) Rs. 790
Solutions
S1. Ans.(a)
Sol. Direct trick-
(9 + 6) : (9 – 6)
15 : 3
S2. Ans.(b)
Sol.                     Sister                Brother
10 years before     3x                         x
Present                (3x + 10)               (x+10)
(3x + 20) = (x + 20) × 2
3x + 20 = 2x + 40
x = 20
Aman = 20 +10 = 30
S3. Ans.(d)
Sol. M=3(S1+S2+S3 )
(M+5)=(S1+5)+(S2+5)+(S3+5)
(M+5)=(S1+S2+S3+15)×2
(M+5)=((M/3)+15)×2
M = 75 years

S5. Ans.(c)
Sol.
Original area = l × b
New area =(5/4) l × (5/4) b = (25/16) lb
Increase % =((25lb/16) – lb) × 100%
= 56.25%
S6. Ans.(c)
Sol. (C.P.)/(S.P.)=25/24
=(1/25)×100
= 4%
S7. Ans.(d)
Sol. 4%=1/25
C.P. = 25
S.P. = 24
SP: CP = 24 : 25
S8. Ans.(a)
Sol. (S.P.1)/(100 ± a)=(S.P.2)/(100 ± b)
(800/80)=(x/120)
x = 1200
S9. Ans.(a)
Sol.
(S.P.1)/(100 ± a)=(S.P.2)/(100 ± b)
(188/94)=(x/106)
x = 212