**Q1. In what time will a 100 metre long train running with a speed of 50 km/hr cross a pillar?**

(a) 7.0 sec

(b) 72 sec

(c) 7.2 sec

(d) 70 sec

**Q3. In a triangle PQR, the side QR is extended to S. ∠QPR = 72° and ∠PRS = 110°, then the value of ∠PQR is:**

(a) 38°

(b) 32°

(c) 25°

(d) 29°

**Q5. In ∆ABC, ∠B = 70° and ∠C = 60°. The internal bisectors of the two smallest angles of ∆ABC meet at O. The angle so formed at O is:**

(a) 125°

(b) 120°

(c) 115°

(d) 110°

**Q6. The mean of range, mode and median of the data 4, 3, 2, 2, 7, 2, 2, 0, 3, 4, 4 is**

(a) 4

(b) 5

(c) 2

(d) 3

**Q7. A cylindrical container of 32 cm height and 18 cm radius is filled with sand. Now all this sand is used to form a conical heap of sand. If the height of the conical heap is 24 cm, what is the radius of its base?**

(a) 12 cm

(b) 24 cm

(c) 36 cm

(d) 32 cm

**Q8. The following table given the monthly income of 10 families in a city: Income (Rs.): 4600, 5560, 6440, 4530, 7670, 6850, 6750, 7910, 5490, 6800. Calculate the arithmetic mean.**

(a) 4897

(b) 6542

(c) 6260

(d) 7260

**Q9. The radius of a cylinder is 7 cm and height is 20 cm. the volume of the cylinder is **

(a) 3080 cm³

(b) 1540 cm³

(c) 2310 cm³

(d) 3850 cm³

**Q10. The ratio of the radius of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is **

(a) 7 : 6

(b) 4 : 9

(c) 20 : 27

(d) 10 : 9

SOLUTIONS

S1. Ans.(c)

Sol.

time = (100 meter)/(50×5/18 m/sec) = 7.2 sec.

S2. Ans.(c)

Sol.

(l + m + n)2 = l2+ m2 + n2 + 2(lm + mn + nl)

81 = 31 + 2 (lm + mn + nl)

(lm + mn + nl) = 25

S3. Ans.(a)

Sol.

∠PQR = (∠PRS – ∠QPR)

= 110° – 72° = 38°

S4. Ans.(a)

Sol.

x+1/x=√3

cubing both sides,

x^3+1/x^3 +3(x+1/x)=3√3

x^3+1/x^3 =3√3-3√3=0

S5. Ans.(a)

Sol

AOC = 90+(∠B)/2

= 90 + 35 = 125°

S6. Ans.(a)

Sol. After arranging the given numbers in ascending order, we get 0, 2, 2, 2, 2, 3, 3, 4, 4, 4, 7

Range = Maximum value – Minimum value = 7 – 0 = 7

Here, the number of terms = 11

The median for an odd number of terms is the middle term when the terms are arranged in ascending or descending order.

When the number of terms ‘n’ is odd, then the middle term is ((n + 1))/2 th term.

∴ Median = ((11 + 1)/2)^th term

= 6th term

= 3

Here, 2 is getting repeated the maximum number of times, i.e., 4 times.

Therefore, the mode = 2

Mean of range, mode and median

=(7+2+3)/3=12/3=4

S7. Ans.(c)

Sol.

Volume of cylinder and cone are equal.

so, πr_1^2 h_1=1/3 πr_2^2 h_2

so, (3×18×18×32)/24=r_2^2

r_2=√((3×18×18×32)/24)=36 cm

S8. Ans.(c)

Sol. Mean =

(Sum of all observations)/(Total number of observations)

=((4600+5560 . . . . . . . . 6800))/10

=62600/10=6260

S9. Ans.(a)

Sol. Volume of cylinder = πr²h

=22/7×7^2×20

= 3080 cm³

S10. Ans.(c)

Sol. Let the radii of two cylinders be 2r and 3r and heights be 5h and 3h respectively.

Volume of first cylinder = π (2r)² (5h) = 20π r²h

Volume of second cylinder = π (3r)² (3h) = 27 π r²h

Required ratio

=(20πr^2 h)/(27πr^2 h)

=20/27=20∶27