**Q1. How many sq. cm of cardboard is required to make a Jewelry box of dimensions 50cm, 40cm and 20cm?**

(a) 4500 sq. cm

(b) 5600 sq. cm

(c) 7600 sq. cm

(d) 8200 sq. cm

**Q2. A wire when bent in the form of circle encloses an area of 154 sq. cm. If the same wire is bent in form of square, what is the area enclosed by it?**

(a) 100 sq. cm

(b) 144 sq. cm

(c) 121 sq. cm

(d) 81 sq. cm

**Q3. The sum of digits of a two-digit number is 14. When the digits are reversed, the number decreases by 18. Find the original number?**

(a) 68

(b) 86

(c) 78

(d) 87

**Q4. Find the LCM of 3/5,16/15 and 7/20.**

(a) 336/5

(b) 1/5

(c) 336/60

(d) 1/60

**Q5. A car travels at a speed of 50 km/h for one-third of the journey, 60 km/hr for next one-third of the journey and remaining at the speed of 30 km/h. Find the average speed of the car for the entire journey?**

(a) 42 6/7 km/h

(b) 33 4/7 km/h

(c) 41 5/7 km/h

(d) 38 5/7 km/h

**Q6. A positive number exceed its positive square root by 56. Find the number.**

(a) 8

(b) 64

(c) 9

(d) 81

**Q7. Compute 0.99 × 365 =**

(a) 360.85

(b) 361.55

(c) 361.35

(d) 362.55

**Q8. Compute: (21 + 4 × 5.5) ÷ 0.043**

(a) 1

(b) 10

(c) 100

(d) 1000

**Q9. The mean of median and mode of the data 7, 6, 7, 9, 8, 8, 10, 8 is**

(a) 5.5

(b) 8

(c) 8.5

(d) 9

**Q10. Two poles of height 13 m and 18 m stand vertically upright on a plane ground. If the distance between their feet is 12 m. Find the distance between their tops.**

(a) 13 m

(b) 17 m

(c) 15 m

(d) 14 m

SOLUTIONS

S1. Ans.(c)

Sol

Required area of cardboard = 2(50×40+40×20+20×50)=2(2000+800+1000)=2×3800

= 7600 sq. cm

S2. Ans.(c)

Sol.

22/7×r^2=154

r^2=(154×7)/22

r = 7 cm

ATQ,

2πr = 4a

2×22/7×7=4a

a= 11 cm

Area of square = a² = 121 sq. cm

S3. Ans.(b)

Sol.

Let two digit no be 10x + y

ATQ,

x + y = 14 ….(i)

10x + y – (10y + x) = 18

9x – 9y = 18

x – y= 2 …(ii)

Solving (i) & (ii)

x = 8, y = 6

Original no. = 10 × 8 + 6 = 86

S4. Ans.(a)

Sol.

L.C.M of[3/5,16/15,7/20]=(L.C.M of (3,16,7))/(H.C.F of (5,15,20) )=336/5

S5. Ans.(a)

Sol.

Let total distance be 3x

Average speed =3x/(x/50+x/60+x/30)

=3/((6+5+10)/300)

=(3×300)/21=42 6/7 km/h

S6. Ans.(b)

Sol.

Let no. be x²

x² – x = 56

x [x – 1] = 56

x [x – 1] = 8 × 7

x = 8

Number is x² = 64

S7. Ans.(c)

Sol.

99/100×365

((100-1)×365)/100=365-3.65

= 361.35

S8. Ans.(d)

Sol.

(21 + 22) × 1000/43

=43×1000/43=1000

S9. Ans.(b)

Sol. After arranging 7, 6, 7, 9, 8, 8, 10, 8 in ascending order, we get 6, 7, 7, 8, 8, 8, 9, 10

Here, 6 is once, 7 is twice, 8 is thrice, 9 is once and 10 is once.

As 8 occurs the maximum number of times, i.e., three times, the mode is 8.

Here, the number of terms, n = 8

As the number of terms is an even number,

Median = 1/2

[(n/2)th term+(n/2+1)th term]

=1/2 [4th term+(4+1)th term]

=1/2 [4th term+5th term]

=18 [8+8]

= 8

∴ Mean of median and mode = (8 + 8)/2=8

S10. Ans.(a)

Sol.

In rt. ∆ABE,

AE² = (5)² + (12)²

AE = √(25+144)=13m