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Maths Quiz for KVS/NVS and CTET Exams

Maths Quiz for KVS/NVS and CTET Exams_30.1


Q1. The average of marks obtained by 120 candidates was 35. If the average of marks of passed candidates was 39 and that of failed candidates was 15, the number of candidates who passed the examination is:
(a) 100
(b) 110
(c) 80
(d) 70

Q2. In a school with 600 students, the average age of the boys is 12 years and that of the girls is 11 years. If the average age of the school is 11 years and 9 months, then the number of girls in the school is:
(a) 450
(b) 150
(c) 250
(d) 350

Q3. The average expenditure of a man for the first five months is Rs. 3600 and for next seven months it is Rs. 3900. If he saves Rs. 8700 during the year, his average income per month is:
(a) Rs. 4500
(b) Rs. 4200
(c) Rs. 4050
(d) Rs. 3750

Q4. The average of two numbers A and B is 20, that of B and C is 19 and of C and A is 21. What is the value of A?
(a) 24
(b) 22
(c) 20
(d) 18

Q5. 12 men can finish a project in 20 days. 18 women can finish the same project in 16 days and 24 children can finish it in 18 days. 8 women and 16 children worked for 9 days and then left. In how many days will 10 men complete the remaining project?
(a) 10 1/2
(b) 10
(c) 9
(d) 11 1/2
           
Q6. An item was bought at Rs. X and sold at Rs. Y, thereby earning a profit of 20%. Had the value of X been 15% less and the value of Y been Rs. 76 less, a profit of 30% would have been earned what was the value of X?
(a) Rs. 640
(b) Rs. 400
(c) Rs. 600
(d) Rs. 800

Directions (7-10): In the following questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
(a) If x ≥ y
(b) If x ≤ y
(c) If x < y
(d)If relationship between x and y cannot be established

Q7. I. x2-3x-88=0
II. y2+8y-48=0

Q8. I. 5×2+29x+20=0
II. 25y2+25y+6=0

Q9.I. 2×2-11x+12=0
II. 2y2-19y+44=0

Q10.I. 3×2+10x+8=0
II. 3y2+7y+4=0

Solutions

S1. Ans.(a)
Sol. Let, the number of candidates who passed = x
Then, 39×x+15×(120-x)=120×35
∴ 24x = 4200 – 1800
Or, x=2400/24
x = 100.

S2. Ans.(b)
Sol. Number of girls = x
Number of boys = 600 – x
∴ (600 – x) × 12 + 11x
=11 ¾ × 600=47/4×600
⇒ 7200 – 12x + 11x = 7050
⇒ x = 7200 – 7050 = 150

S3. Ans.(a)
Sol. Total expenditure for the first five months = 5 × 3600 = Rs. 18000
Total expenditure for the next seven months = 7 × 3900 = Rs. 27300
Savings = Rs. 8700
Total income during the year
= 18000 + 27300 + 8700 = Rs. 54000
Average income per month =54000/12 = Rs. 4500.

S4. Ans.(b)
Sol. According to the question,
(A + B)/2=20
⇒ A + B = 40 …(i)
(B + C)/2=19
⇒ B + C = 38  …(ii)
(C + A)/2=21
⇒ C + A = 42 …(iii)
Adding (i), (ii) and (iii), we get
∴ 2A + 2B + 2C = 120
⇒ A + B + C = 60
∴ A = (A + B + C) – (B + C)  …(iv)
= 60 – 38 = 22

S5. Ans.(b)
Sol. 12M × 20 = 18W × 16 = 24C × 18
5M : 6W : 9C
Work done by 8 women and 16 children
(8W+16C)×9=(8×5/6 M+16×5M/9)×9
=(40M/6+80M/9)×9=140 M
Remaining work = 12M × 20 – 140M = 240M – 140M = 100M
10 men will complete the work in 100/10 days = 10 days

S6. Ans.(d)
Sol. Given, the CP of an item = Rs. X
And SP of an item = Rs. Y
Here, Y = 1.2X
If the CP of an item be 15% less = 85% of X = 0.85X
According to the question,
0.85 X × 130/100=1.2X-76
⇒ 11.05X = 12X – 760
⇒ 12X – 11.05X = 760
⇒ 0.95X = 760
⇒ X= 760/0.95
⇒ X = 800
∴ CP of an item is Rs. 800.

S7. Ans.(d)
Sol. x2-3x-88=0
⇒ x2-11x+8x-88=0
⇒ x(x-11)+8(x-11)=0
⇒ (x+8)(x-11)=0
⇒ x = -8 or 11
II. y2+8y-48=0
⇒ y2+12y-4y-48=0
⇒ y(y+12)-4(y+12)=0
⇒ (y-4)(y+12)=0
⇒ y = -12 or 4
Hence, it cannot be established.

S8. Ans.(c)
Sol. I. 5×2+29x+20=0
⇒ 5×2+25x+4x+20=0
⇒ 5x(x+5)+4(x+5)=0
⇒ (5x+4)(x+5)=0
⇒ x=-5 or (-4)/5
II. 25y2+25y+6=0
⇒ 25y2+15y+10y+6=0
⇒ 5y(5y+3)+2(5y+3)=0
⇒ (5y+2)(5y+3)=0
⇒ y=(-3)/5 or (-2)/5
Hence, x < y

S9. Ans.(b)
Sol. I. 2×2-11x+12=0
⇒ 2×2-8x-3x+12=0
⇒ 2x(x-4)-3(x-4)=0
⇒ (2x-3)(x-4)=0
⇒ x=4 or 3/2
II. 2y2-19y+44=0
⇒ 2y2-8y-11y+44=0
⇒ 2y(y-4)-11(y-4)=0
⇒ (y-4)(2y-11)=0
⇒ y = 4 or 11/2
Hence, x ≤ y

S10. Ans.(b)
Sol. I. 3×2+10x+8=0
⇒ 3×2+6x+4x+8=0
⇒ 3x(x+2)+4(x+2)=0
⇒ (3x+4)(x+2)=0
⇒ x = -2 or (-4)/3
II. 3y2+7y+4=0
⇒3y2+3y+4y+4=0
⇒ 3y(y+1)+4(y+1)=0
⇒ (y+1)(3y+4)=0
⇒ y=-1 or (-4)/3
Hence, x ≤ y