Question 1 :
The in circle of triangle ABC touches BC at D, AC at E, and AB at F. Given AF = 3, BD = 4, and CE = 6, find the perimeter of triangle ABC
(A) 26 (B) 55 (C) 37
Solution :
AF = AE = 3, BD = BF = 4 and CE = CD = 6
Perimeter of the triangle ABC = AB + BC + CA
AB = AF + FB = 3 + 4 ==> 7
BC = BD + DC = 4 + 6 ==> 10
CA = CE + EA = 6 + 3 ==> 9
Perimeter of the triangle = 7 + 10 + 9
= 26 cm
Question 2 :
Find the sum of the first 16 terms of the arithmetic series 5, 10, 15 . . .
(A) 490 (B) 230 (C) 510
Solution :
The sum of n terms = sn = (n/2)[2a + (n - 1)d]
a = 5, d = t2 - t1 = 10 - 5 ==> 5
s16 = (16/2)[2(5) + (16 - 1)(5)]
s16 = 8[10 + (15)5]
s16 = 8[10 + 75]
s16 = 8[85]
= 425
Question 3 :
Subtract 573.9246 - 215.6
(A) 320.1046 (B) 125.3246 (C) 358.3246
Solution :
573.9246 - 215.6 = 358.3246
Question 4 :
Calculate 2 × 0 × 0 × 8 x 12 =
(A) 5 (B) 0 (A) 3
Solution :
In the expression, one of the number is 0. Hence the answer is 0.
Question 5 :
Express 11/5 in decimal form
(A) 5.5 (B) 2.2 (C) 8.2
Solution :
By dividing 11 by 5, we get 2.2
Question 6 :
The sum of the angles of a triangle is
(A) 180 degree (B) 240 degree (C) 360 degree
Solution :
The sum of the interior of the triangle is 180 degree.
Question 7 :
The circle is having 14 cm as diameter. Then what will be the perimeter of the circle
(A) 25π (B) 50π (C) 14π
Solution :
Diameter of the circle = 14 cm
radius of the circle = 7 cm
Perimeter of the circle = Circumference of circle
= 2πr
= 2π(7)
= 14π
Question 8 :
Divide 0.4096 ÷ 0.032
(A) 12.8 (B) 15.3 (C) 10.1
Solution :
0.4096 ÷ 0.032 = 0.4096 / 0.032
Multiplying the numerator and denominator by 1000, we get
= 4096/320
= 12.8
Question 9 :
Reduce 45% into a fraction.
(A) 9/20 (B) 9/17 (C) 5/19
Solution :
45% = 45/100
Dividing the numerator and denominator by 5 times table. we get
= 9/20
Question 10 :
Multiply 6a x 2a
(A) 8a (B) 12a2 (C) 8a2
Solution :
6a x 2a = 12a2
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