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Maths Quiz for CTET Exam 2016

1. X can do a piece of work in 20 days and Y can do it in 10 days. They started together, but after 6 days X leaves off. Y will do the rest work in
(a) 1 day
(b) 2 days
(c) 7/3 days
(d) 3 days
Sol.
(X + Y)’s 6 days’ work = 6 * (1/20 + 1/10) = 9/10
Remaining work = 1 – 9/10 = 1/10
1/10 work is done 2y Y in 10 * 1/10 = 1 day

2. If each side of a rectangle is increased by 50%, its area will increase by
(a) 125%
(b) 75%
(c) 100%
(d) 225%
Sol.
Original area = l × b
New area =3/2 l×3/2 b=9/4 lb
Increase % =(9/4 lb – lb)/lb×100%
= 125%

3. Ratio of angles of linear pair is 8 : 1 then measure of each angle is
(a) 80°, 10°
(b) 120°, 15°
(c) 160°, 20°
(d) 200°, 25°
Sol.
Let angles be 8x and x
8x + x = 180
x = 20° then angles are 160° and 20°

4.60% of that the students in a school are boys. If the number of girls in the school is 300, then the number of boys is-
(a) 300
(b) 450
(c) 500
(d) 750
Sol:
Here 40% are girl in a school.
40%=300
60%=300× 60/40
=450

5. The circumference of the base of a right circular cylinder is 44 cm and its height is 15 cm. The volume (〖in cm〗^3 ) of the cylinder is (use π=22/7)
(a) 770
(b) 1155
(c) 1540
(d) 2310
Sol.
Radius of cylinder =44/2π=44/(2 × 22/7)=7 cm
∴ Volume of cylinder
=πr^2 h=22/7×7×7×15=2310 〖cm〗^3

6. In a polyhedron, if the number of faces, edges and vertices are denoted by F, E and V respectively, then according to Euler’s formula
(a) F+V-E=2
(b) F-V+E=2
(c) F-V-E=2
(d) F+V=E-2
Sol.
In, Polygon the number of faces, edges and vertices are denoted by F,E and V respectively. According to Euler’s formul a.
(F+V-E=2)

7. The ratio of the length to the breadth of a rectangle is 2 : 1. Its perimeter is 60 cm, find its area.
(a) 250 sq cm
(b) 200 sq cm
(c) 300 sq cm
(d) 205 sq cm
Sol.
2(2x + x) = 60 cm
⇒ x = 10 cm
∴ Area = 2x
x = 2(100) = 200 cm2

8. If x^2+1/x^2 =34, then find the value of x+1/x
(a) 4
(b) 6
(c) 8
(d) 17
Sol.
x^2+1/x^2 =34
⇒ (x+1/x)^2-2×1/x=34
[a^2+b^2=(a+b)^2=2ab]
⇒ (x+1/x)^2=36
⇒ x+1/x=6

9. The value of 5-(2 1/2-3/4)+(3 1/2-1 1/4) is
(a) 4 1/2
(b) 5 1/2
(c) 5 1/4
(d) 3 1/2
Sol.
5-[5/2-3/4]+[7/2-5/4]=5-[(10 –  3)/4]+[(14 –  5)/4]=5-7/4+9/4
=(20 – 7 + 9 )/4=22/4=11/2=5 1/2

10. One-fourth of a pizza was eaten by Renu. The rest was equally distributed among 12 children. What part of the pizza did each of these children get?
(a) 1/16
(b) 1/32
(c) 3/16
(d) 1/8
Sol.
Pizza remaining = 3/4
Pizza that single child get = (3/4)  × (1/12)=1/16 Join India's largest learning destination

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