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# Maths Questions for DSSSB Exam 2017 Q1. Find the length of the longest rod that can be placed in a half of 10 m length 6 m breadth and 4 m height,
(a) 2√38 m
(b) 4√38 m
(c) 2√19 m
(d) √152 m

Q2. The difference of the areas of two squares drawn on two line segments of different lengths is 32sqcm. Find the length of the greater line segment if one is longer than the other by 2 cm.
(a) 7 cm
(b) 9 cm
(c) 11 cm
(d) 16 cm

Q3. A took 15 sec. to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68 m/min. The area of the field is:
(a) 30 m^2
(b) 40 m^2
(c) 50 m^2
(d) 60 m^2

Q4. The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is
(a) 1520 m^2
(b) 2420 m^2
(c) 2480 m^2
(d) 2520 m^2

Q5. The area (in m^2) of the square which has the same perimeter as a rectangle whose length is 48 m and is 3 times its breadth is:
(a) 1000
(b) 1024
(c) 1600
(d) 1042

Q6. On mixing two classes A and B of students having average marks 25 and 40 respectively. The over all average obtained is 30. Find the ratio of the students in the class A and B.
(a) 2 : 1
(b) 5 : 8
(c) 5 : 6
(d) 3 : 4

Q7. At present the ratio of the age of Maya and Chhaya is 6 : 5 and fifteen years from now, the ratio will get changed to 9 : 8. Maya’s present age is
(a) 21 years
(b) 24 years
(c) 30 years
(d) 40 years

Q8. What must be added to each term of the ratio 7 : 11, so as to make it equal to 3 : 4?
(a) 8
(b) 7.5
(c) 6.5
(d) 5

Q9. The sum of three numbers is 68. If the ratio of the first to the second be 2 : 3 and that of the second to the third be 5 : 3, then the second number is
(a) 30
(b) 58
(c) 20
(d) 48

Q10. Two number are in the ratio 4 : 5 and their L.C.M. is 180. The smaller number is
(a) 9
(b) 15
(c) 36
(d) 45
Solutions:
S1. Ans.(d)
Sol. l = 10 m, b = 6 m, h = 4 m
Length of diagonal (longest + rod) = √(100+36+16)=√152 m
S4. Ans.(d)
Sol. Let the breadth be = x m
∴ length = (23 + x) m
⇒ 2(x + 23 + x) = 206
4x = 206 – 46
x = 160/4 = 40 m
∴ length = 40 + 23 = 63 m
∴ Required area = 63 × 40
= 2520 m^2
S5. Ans.(b)
Sol. Length of rectangle = 48 m
Breadth of rectangle = 16 m
According to question,
Perimeter of square =
Perimeter of rectangle
= 2 (48 + 16)
4 × side = 2 × 64
Side = (2 × 64)/4 = 32 m
∴ Area of the square
= (side)^2=(32)^2
= 1024
S8. Ans.(d)
Sol. A/B=7/11 Given
Let x be added to both A & B
⇒(7+x)/(11+x)=3/4
Cross multiply the equation
28 + 4x = 33 + 3x
x = 5
S9. Ans.(a)
Sol. A + B + C = 68
A : B : C
2 : 3
5 : 3
_______________
▁ (10∶15∶9)
_____________
10x + 15x + 9x = 34x
34x = 68
x = 2
∴ A = 2 × 10 = 20
B = 2 × 15 = 30
C = 9 × 2 = 18
S10. Ans.(c)
Sol.
A : B
4 : 5
4x : 5x
LCM = 4 × 5 × x = 20x
20x = 180
x = 9
∴ Smallest number is
= 4 × 9 = 36
Largest number is = 5 × 9 = 45