AP NMMS 2024 Previous Year Question Paper with Answers | NMMS Andhra Pradesh Key & Solutions

PART-2
AP NMMS 2024
SAT (Scholastic Aptitude Test)
Mathematics

126) The sum of reciprocals of two numbers is $\frac{1}{4}$ .How many times their sum is equal to their product?

A) 5
B) 4
C) 7
D) 10

View Answer

B) 4

Explanation:Solution:
Given $\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{1}{4}$ .
So $\displaystyle \frac{a+b}{ab}=\frac{1}{4}\Rightarrow ab=4(a+b)$ .
Thus the product equals $4$ times the sum.
Answer: (2) 4

127) How many edges does the cube having 6 faces and 8 vertices?

A) 6
B) 8
C) 12
D) 14

View Answer

C) 12

Explanation:Solution:
A cube has 12 edges.
Answer: (3) 12

128) The simplified value of $\left(\frac{x^7}{x^9}\right)^{\frac{1}{63}} \times \left(\frac{x^9}{x^3}\right)^{\frac{1}{27}} \times \left(\frac{x^3}{x^7}\right)^{\frac{1}{21}}$ is

A) 0
B) 35721
C) 9
D) 1

View Answer

D) 1

Explanation:Solution:
Expression = $\left(\frac{x^7}{x^9}\right)^{1/63}\times\left(\frac{x^9}{x^3}\right)^{1/27}\times\left(\frac{x^3}{x^7}\right)^{1/21}$ .
Simplify exponents:
$\left(x^{-2}\right)^{1/63}=x^{-2/63},\quad x^{6/27}=x^{2/9},\quad x^{-4/21}$ .
Sum exponents: $-\tfrac{2}{63}+\tfrac{2}{9}-\tfrac{4}{21}=0$ .
So expression $=x^0=1$ .
Answer: (4) 1

129) If two cubes each of edge 4 cm are placed end to end, then the dimensions of the resulting solid are

A) 8 cm, 4 cm, 4 cm
B) 4 cm, 4 cm, 4 cm
C) 6 cm, 4 cm, 4 cm
D) 8 cm, 8 cm, 4 cm

View Answer

A) 8 cm, 4 cm, 4 cm

Explanation:Solution:
Two cubes of edge $4$ cm placed end-to-end give solid with dimensions $8$ cm, $4$ cm, $4$ cm.
Answer: (1) $8$ cm, $4$ cm, $4$ cm

130) A cricketer scores the following runs in eight innings:
28, 59, 30, 100, 45, 0, 76, 99
Find the range of his scores.

A) 99
B) 0
C) 100
D) 59

View Answer

C) 100

Explanation:Solution:
Scores: $28,59,30,100,45,0,76,99$ .
Range $=$ max – min $=100-0=100$ .
Answer: (3) 100

131) The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

A) 68 cm
B) 60 cm
C) 480 cm
D) 46 cm

View Answer

A) 68 cm

Explanation:Solution:
Diagonals of rhombus are $16$ cm and $30$ cm. Side $=\sqrt{(16/2)^2+(30/2)^2}=\sqrt{8^2+15^2}=\sqrt{289}=17$ cm.
Perimeter $=4\times17=68$ cm.
Answer: (1) 68 cm

132) If A, B, C represent 3 semicircles, then the circumference of the whole diagram is

A) 22 cm
B) 66 cm
C) 132 cm
D) 44 cm

View Answer

B) 66 cm

Explanation:Solution:
Diameters of semicircles (from diagram) are $7$ cm, $14$ cm and $21$ cm (the big one equals $7+14$ ).
Total outer curved length = sum of semicircle circumferences
$=\tfrac{\pi}{2}(7+14+21)=\tfrac{\pi}{2}\times42=21\pi$ .
Using $\pi=\tfrac{22}{7}$ : $21\pi=21\times\tfrac{22}{7}=66$ cm.
Answer: (2) 66 cm

133) What is the probability of ‘M’ in the word MATHEMATICS?

A) $\frac{1}{11}$
B) $\frac{4}{11}$
C) $\frac{7}{11}$
D) $\frac{2}{11}$

View Answer

D) $\frac{2}{11}$

Explanation:Solution:
Word “MATHEMATICS” has 11 letters; number of ‘M’ = 2.
Probability $=2/11$ .
Answer: (4) $\tfrac{2}{11}$

134) Which of the following measurements are not the sides of a triangle?

A) 8 cm, 9 cm, 10 cm
B) 3 cm, 4 cm, 5 cm
C) 1 cm, 2 cm, 3 cm
D) 5 cm, 4 cm, 6 cm

View Answer

C) 1 cm, 2 cm, 3 cm

Explanation:Solution:
Triangle inequality: three lengths form a triangle only if sum of any two > third.
Option (3) $1,2,3$ fails because $1+2=3$ (degenerate), so not a triangle.
Answer: (3) $1\text{ cm},2\text{ cm},3\text{ cm}$

135) In the given parallelogram, what is the value of x?

A) 13
B) 3
C) 16
D) 7

View Answer

B) 3

Explanation:Solution:
In a parallelogram diagonals bisect each other ⇒ opposite segments of a diagonal are equal.
From diagram: the segment opposite $20$ is $y+7$ ⇒ $y+7=20$ ⇒ $y=13$ .
The segment opposite $16$ is $x+y$ ⇒ $x+y=16$ . Substitute $y=13$ ⇒ $x=3$ .
Answer: (2) $3$

136) If the multiplicative inverse of a number is the number itself, then the number is/are

A) 1 and –1
B) 1 only
C) –1 only
D) 0

View Answer

A) 1 and –1

Explanation:Solution:
If multiplicative inverse equals the number: $1/x=x\Rightarrow x^2=1\Rightarrow x=\pm1$ .
So both $1$ and $-1$ .
Answer: (1) $1$ and $-1$

137) The value of $\frac{\sqrt[3]{8}}{\sqrt{16}} ÷ \sqrt{\frac{100}{49}} \times \sqrt[3]{125}$ is

A) $\frac{7}{100}$
B) $\frac{7}{4}$
C) $\frac{4}{7}$
D) 1

View Answer

B) $\frac{7}{4}$

Explanation:Solution:
Interpretation: $\dfrac{\sqrt[3]{8}}{\sqrt{16}}\div\sqrt{\dfrac{100}{49}}\times\sqrt[3]{125}$ .
Compute: $\sqrt[3]{8}=2,\ \sqrt{16}=4\Rightarrow 2/4=\tfrac{1}{2}.$
$\sqrt{100/49}=10/7,\ \sqrt[3]{125}=5.$
So value $=\tfrac{1}{2}\div\frac{10}{7}\times5=\tfrac{1}{2}\times\frac{7}{10}\times5=\tfrac{7}{4}.$
Answer: (2) $\tfrac{7}{4}$

138) Each side of a square is 10 m long.A rectangle of the same perimeter has sides in the ratio 2 : 3.What is the area of the rectangle in square meters?

A) 96 sq.m
B) 106 sq.m
C) 86 sq.m
D) 384 sq.m

View Answer

A) 96 sq.m

Explanation:Solution:
Square side $=10$ m ⇒ perimeter $=40$ m.
Rectangle with same perimeter and side ratio $2:3$ : let sides $2k$ and $3k$ .
Perimeter $=2(2k+3k)=10k=40\Rightarrow k=4$ . Sides $=8$ m and $12$ m.
Area $=8\times12=96$ sq. m.
Answer: (1) 96 sq. mt.

139) A car covers a distance of 76.5 km in 1.7 hours.What is the average distance covered by it in 1 hour?

A) 17 km
B) 35 km
C) 48 km
D) 45 km

View Answer

D) 45 km

Explanation:Solution:
Average distance per hour $=76.5\text{ km} \div 1.7\text{ h}=45$ km/h.
Answer: (4) 45 km

140) Match the following

A) i-(a), ii-(b), iii-(c), iv-(d)
B) i-(c), ii-(a), iii-(d), iv-(b)
C) i-(a), ii-(d), iii-(b), iv-(c)
D) i-(a), ii-(d), iii-(c), iv-(b)

View Answer

B) i-(c), ii-(a), iii-(d), iv-(b)

Explanation:Solution (matching line symmetries):
i) square ⇒ 4 lines (c)
ii) equilateral triangle ⇒ 3 lines (a)
iii) rectangle with unequal adjacent sides? (diagram shows two equal marks vertically) ⇒ 2 lines (d)
iv) letter “Z” ⇒ 0 lines (b)
So matching: i-(c), ii-(a), iii-(d), iv-(b) ⇒ option (2).
Answer: (2) i-(c), ii-(a), iii-(d), iv-(b)

141) If $3 \times 3 \times 3 \times \dots , n \text{ times} = 2187$ , then what is the value of (n)?

A) 7
B) 6
C) 2
D) 3

View Answer

A) 7

Explanation:Solution:
$3\times3\times\cdots$ (n times) $=3^n=2187$ . Since $3^7=2187$ , $n=7$ .
Answer: (1) 7

142) The number of girls in class are 20, which are equal to 40% of the class.Then the total number of students in class is

A) 80
B) 40
C) 60
D) 50

View Answer

D) 50

Explanation:Solution:
Girls $=20$ are $40\%$ of class ⇒ total $=20/0.4=50$ .
Answer: (4) 50

143) If $(525)^2 = 275625$ , then what is the value of $\sqrt{27.5625}$ ?

A) 0.525
B) 52.5
C) 5.25
D) 0.0525

View Answer

C) 5.25

Explanation:Solution:
Given $(525)^2=275625$ . Then $\sqrt{27.5625}=\sqrt{\tfrac{275625}{10000}}=\tfrac{525}{100}=5.25$ .
Answer: (3) 5.25

144) Rani has 3 times as many two-rupee coins as she has five-rupee coins.If she has in all a sum of Rs.77, how many coins of Rs.5 and Rs.2 denomination does she have?

A) 5, 15
B) 8, 21
C) 4, 12
D) 7, 21

View Answer

D) 7, 21

Explanation:Solution:
Let number of Rs.5 coins $=x$ . Then Rs.2 coins $=3x$ .
Total value $=5x+2(3x)=5x+6x=11x=77\Rightarrow x=7$ .
So Rs.5 coins $=7$ , Rs.2 coins $=21$ .
Answer: (4) $7,\ 21$

145) In the $AD \parallel BC$ , then find $\angle BAC$ .

A) 50°
B) 60°
C) 70°
D) 80°

View Answer

B) 60°

Explanation:Solution:
AD ∥ BC. From diagram angle at B (∠ABC) is $85^\circ$ . The small angle between AC and AD (exterior at A) is $35^\circ$ . Because AD ∥ BC, angle between AC and AD equals angle ACB (alternate interior).
So ∠ACB $=35^\circ$ . Now triangle angles sum:
∠BAC $=180^\circ-∠ABC-∠ACB=180^\circ-85^\circ-35^\circ=60^\circ$ .
Answer: (2) $60^\circ$

146) Individual soul is part of the entity.This is the principle of

A) Dwaitha
B) Vishishtadwaitha
C) Adwaitha
D) Dwaithadwaitha

View Answer

B) Vishishtadwaitha

147) East India Company began its trade in which state in India?

A) Tamil Nadu
B) Gujarat
C) Bengal
D) Andhra Pradesh

View Answer

C) Bengal

148) Who among the following was/were associated to Ryotwari System?

A) Captain Alexander Read
B) Thomas Munro
C) Both (1) and (2)
D) None of the two

View Answer

C) Both (1) and (2)

149) Correct chronological order of Dynasties

A) Mamluk – Tughlaq – Khilji – Sayyid – Lodi
B) Tughlaq – Mamluk – Sayyid – Khilji – Lodi
C) Mamluk – Khilji – Tughlaq – Sayyid – Lodi
D) Lodi – Sayyid – Mamluk – Khilji – Tughlaq

View Answer

C) Mamluk – Khilji – Tughlaq – Sayyid – Lodi

150) The tillers had to pay ______ of their produce as land revenue in the reign of Delhi Sultanate.

A) ½
B) ⅓
C) ⅔
D) ¼

View Answer

B) ⅓

151) The founder of Kalyani Chalukyas

A) Bhillama
B) Tailapa II
C) Bettiga Vishnuvardhan
D) Kulasekhara

View Answer

B) Tailapa II

152) Identify the wrongly matched pair.

A) Prola II – The first independent ruler of Kakatiyas
B) Mahadeva – Died while besieging the Devagiri fort
C) Pratapa Rudra – Constructed Rudreswara temple at Hanumakonda
D) Ganapathi Deva – His period is called the Golden era

View Answer

C) Pratapa Rudra – Constructed Rudreswara temple at Hanumakonda

153) ‘Nrutta Ratnavali’ was written by

A) Jayapa Senani
B) Palkuriki Somanatha
C) Nannechoda
D) Vidhyanadha

View Answer

A) Jayapa Senani

154) Which of the following aspects is not related to Sri Krishna Deva Raya?

A) Scholars known as Ashtadiggajas
B) Telugu work named ‘Amuktamalyada’
C) Title of ‘Andhra Bhoja’
D) Friendly relations with the French and the British

View Answer

D) Friendly relations with the French and the British

155) It is said that even if he was the emperor of India, he would cover his own expenses with the money he earned by sewing hats.

A) Akbar
B) Shajahan
C) Aurangzeb
D) Sher Shah

View Answer

C) Aurangzeb

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PART-2 AP NMMS 2024 SAT (Scholastic Aptitude Test) Mathematics

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PART-2
AP NMMS 2024
SAT (Scholastic Aptitude Test)
Mathematics