Maths quiz (Paper II) for CTET exam_00.1
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Maths quiz (Paper II) for CTET exam

Maths quiz (Paper II) for CTET exam_40.1


1. Rahul bought 2 old scooters for rupees 22,000. By selling one at Profit of 10% and other at loss of 12%, he neither gains nor loses. What is the C.P of the 2 scooters, individually?
A. 12,000; 10,000
B. 13,000; 9000
C. 11,000; 11,000
D. 10,500; 11,500
(Let C.P=y, S.P=22,000(as neither gain nor loss)
 y × 110/100 +(22,000-y)× 88/100 =22,000
11y/10 + 1936000-88y/100 = 22,000
22y = 2,64,000 ; y= 12,000
C.P= 12,000 and 10,000)

2. A shopkeeper loses the S.P of 4 pencils on selling 36 pencils. His loss% –
A. 12.5%
B. 10%
C. 15%
D. 25%
 (Let S.P of 1 pencil= 1 Rs , S.P of 36 pencils= 36 Rs
 S.P<C.P (as loss) , C.P= 36+4= 40 Rs.
 L%= (4/40) ×100= 10% )

3. If a, b, c are respectively the number of faces, edges and vertices of a pentagonal pyramid, then the value of (a-b+c/2)2 – 2 is-

A. 2
B. 1.75
C. -1
D. -1.5
(Euler’s rule: F+V-E=2, a+c–b = 2
By putting it in question: (2/2)2 – 2 = -1 ans)


4. An amount of Rs. 15,000 is distributed among A, B and C in the ratio 4:5:6. What is the share of B?
A. 5000
B. 6000
C. 4000
D. 7000
(let, share of A=4x, B=5x, C=6x
Total amount = A+B+C ; 4x+5x+6x= 15000
15x= 15000, x= 1000
 B’s share(5x) = 5000 Rs)

5. Perimeter of a floor of a cuboidal room is 250m. Cost of painting 4 walls at the rate of 10 rupees per square meter is Rs. 15000. What is the height of the room?
A. 4m
B. 5m
C. 6m
D. 7m
(Area of 4 walls of a room= 2h(l+b)
15000/10= h×2(l+b)
1500= h×250; h=6m)

6. The value of ratio of volume of a sphere to the volume of a cylinder is 2:1 .Find the height of a cylinder-
A. (1/2)r
B. (6/5)r
C. (3/2 r
D. (2/3)r
(Vol. of sphere/Vol. of cylinder) = (2/1)
(4/3)πr3/ πr2h= 2/1

(4r/3h)=(2/1); h= (2/3)r )
7. The square of 9 is divided by the cube root of 125. The remainder is- 
A. 4
B. 1
C. 2
D. 3
( i.e 81/5… remainder is 1)

8. One half of 1.2×10^30 is-
A. 0.6×530
B. 6.0×1030
C. 6.0×1029
D. 1.2×1015
((1.2×1030)/2= 6.0×1029)

9. The number of vertices in a polyhedron which has 30 edges and 12 faces is-
A. 24
B. 12
C. 15
D. 20
(F+V-E=2; 12+V-30=2
V-18=2; V=20)

10. When half of a number is increased by 16, the result is 48. The sum of digits of the original number is-
A. 10
B. 11
C. 12
D. 13
(Let the number be x.
(x/2)+16=48
X+32=96; x=64
Sum= 6+4=10)

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