TS NMMS 2023 Previous Year Question Paper with Answers | NMMS Telangana Key & Solutions


PART-2
MENTAL ABILITY TEST
TS NMMS 2023
SAT (Scholastic Test)
Mathematics

126) The value of \frac{2^{2021}+2^{2019}}{2^{2020}-2^{2018}} is…….

A) 2020
B) \frac{10}3
C) 2^{1000}+1
D) 10

View Answer
D) 10

127) The cost of 7\frac23 meters of rope is ₹12\frac34 then the cost of rope of 1 meter is……

A) ₹2\frac{30}{47}
B) ₹3\frac{18}{25}
C) ₹1\frac{61}{92}
D) ₹6\frac{16}{90}

View Answer
C) ₹1\frac{61}{92}

Problem: 7\tfrac{2}{3}=\dfrac{23}{3} metres of rope cost ₹12\tfrac{3}{4}=\dfrac{51}{4}. Find cost of 1 m.
Solution:
Cost per metre = \dfrac{51/4}{23/3}=\dfrac{51}{4}\cdot\dfrac{3}{23}=\dfrac{153}{92}.
Write as mixed number: \dfrac{153}{92}=1\;\dfrac{61}{92}.
Answer: ₹1\tfrac{61}{92}. (Option 3)

128) The sides of a triangle are in the ratio of \frac12:\frac13:\frac14 If it’s perimeter is 104 cm, then the length of the longest side (in cm) is ……

A) 48
B) 32
C) 26
D) 52

View Answer
A) 48

Explanation:Problem: Sides are proportional to \tfrac12:\tfrac13:\tfrac14. Perimeter =104. Find longest side.
Solution:
Multiply ratios by 12 to clear denominators: sides ∝ 6:4:3.
Sum = 6+4+3=13 parts. One part =104/13=8.
Longest side =6\times8=48 cm.
Answer: 48 cm. (Option 1)

129) The capacity of a tank of dimensions 8m × 6m × 2.5m is ….

A) 120 liters
B) 1200 liters
C) 12000 liters
D) 120000 liters

View Answer
D) 120000 liters

Explanation:Problem: Tank dimensions 8\text{ m}\times6\text{ m}\times2.5\text{ m}. Find capacity in litres.
Solution:
Volume =8\times6\times2.5=120 cubic metres.
1\ \text{m}^3=1000 litres, so capacity =120\times1000=120000 litres.
Answer: 120000 litres. (Option 4)

130) In the given figure, ∠PTQ = 30°, ∠TPQ=34° and ∠SRQ = 45°, then the value of x is……
NMMS 2023 TS

A) 64°
B) 75°
C) 109°
D) 79°

View Answer
C) 109°

Explanation:Problem (angles): In triangle \triangle PTQ, \angle P=34^\circ,\ \angle TPQ=30^\circ. Point S lies on segment TQ. Ray SR meets the baseline at R so that \angle SRQ=45^\circ. Find x, the angle between ST and SR.
Solution:
First find \angle PQT (angle at Q of \triangle PTQ):
\angle Q = 180^\circ - (34^\circ + 30^\circ)=116^\circ.
The baseline is a straight line P\!-\!Q\!-\!R. The angle between TQ and the ray QR equals
\angle TQR = 180^\circ - \angle PQT = 180^\circ - 116^\circ = 64^\circ.
In triangle \triangle SRQ the angles are \angle SQR=64^\circ and \angle SRQ=45^\circ, so the angle at S (between SR and SQ) is
\angle RSQ = 180^\circ - (64^\circ+45^\circ)=71^\circ.
But ST is opposite in direction to SQ (since S lies on segment TQ), so the required angle between ST and SR is the supplement:
x = 180^\circ - 71^\circ = 109^\circ.
Answer: 109^\circ. (Option 3)

131) The smallest natural number to be added to 803642 to get a number which is completely divisible by 9 is……

A) 1
B) 2
C) 3
D) 4

View Answer
D) 4

Explanation:Problem: Smallest natural number to add to 803642 to make it divisible by 9.
Solution:
Sum of digits =8+0+3+6+4+2=23. Nearest multiple of 9 greater than 23 is 27.
Needed addition =27-23=4.
Answer: 4. (Option 4)

132) If \frac x3-\frac{2x}5=\frac{2x}3-\frac{11}{30}, then the value of ‘x’ is……

A) \frac12
B) -\frac43
C) \frac34
D) -\frac34

View Answer
C) \frac34

133) Standard form of 0.00000000234 is……

A) 2.34\times10^{-8}
B) 2.34\times10^{-9}
C) 234\times10^{-10}
D) 23.4\times10^{-11}

View Answer
B) 2.34\times10^{-9}

Explanation:Write 0.00000000234 in standard form.
0.000000001 = 10^{-9}. So 0.00000000234 = 2.34\times10^{-9}.
Answer: 2.34\times10^{-9}. (Option 2)

134) In Δ ABC, AB = BC and in ΔCDE, CD = DE. If ∠BCD = 72°, ∠DEF 124° and ACEF is a straight line, then the value of ‘y’ is……..
NMMS 2023 TS

A) 52°
B) 58°
C) 68°
D) 76°

View Answer
A) 52°

135) The product of \left(x^2\;-\;7y^2\right) and \left(x^2\;+\;4y^2\right) is….

A) x^4\;+\;3x^2y^2\;-\;28y^4
B) x^4\;-\;3x^2y^2\;-\;28y^4
C) x^4\;+\;3x^2y^2\;+\;28y^4
D) x^4\;-\;3x^2y^2\;+\;28y^4

View Answer
B) x^4\;-\;3x^2y^2\;-\;28y^4

Explanation:Compute (x^2-7y^2)(x^2+4y^2).
Multiply: x^2\cdot x^2 + x^2\cdot4y^2 -7y^2\cdot x^2 -7y^2\cdot4y^2
= x^4 +4x^2y^2 -7x^2y^2 -28y^4 = x^4 -3x^2y^2 -28y^4.
Answer: x^4 -3x^2y^2 -28y^4. (Option 2)

136) A vertical stick 2 meter tall casts a shadow 5 meter long. At the same time, if a tree casts a shadow 55 meter long, then the height of the tree (in meters) is……

A) 137.5
B) 27.5
C) 22
D) 10

View Answer
C) 22

Explanation:Similar triangles: stick height 2 m casts shadow 5 m. Tree shadow 55 m, so tree height = \dfrac{2}{5}\times55 = 22 m.
Answer: 22 m. (Option 3)

137) If a number exceeds 20% of itself by 40, then the number is……

A) 50
B) 60
C) 80
D) 320

View Answer
A) 50

Explanation:Let number be n. “Number exceeds 20\% of itself by 40” means n = 0.2n + 40.
So 0.8n = 40 \Rightarrow n = \dfrac{40}{0.8} = 50.
Answer: 50. (Option 1)

138) The value of \sqrt{41+\sqrt{54+\sqrt{88+\sqrt{128+\sqrt{256}}}}} Is…….

A) 7
B) 6
C) 8
D) 10

View Answer
A) 7

Explanation:Evaluate nested radicals:
\sqrt{41+\sqrt{54+\sqrt{88+\sqrt{128+\sqrt{256}}}}}.
Compute inside out:
\sqrt{256}=16.
\sqrt{128+16} = \sqrt{144}=12.
\sqrt{88+12} = \sqrt{100}=10.
\sqrt{54+10} = \sqrt{64}=8.
\sqrt{41+8} = \sqrt{49}=7.
Answer: 7. (Option 1)

139) If two adjacent angles of a parallelogram are 2x°+25° and 3x° – 5°, then the value of ‘x’ is….

A) 28
B) 32
C) 36
D) 42

View Answer
B) 32

Explanation:Two adjacent angles of a parallelogram are 2x^\circ+25^\circ and 3x^\circ-5^\circ.
Adjacent angles in a parallelogram are supplementary, so
(2x+25)+(3x-5)=180 \Rightarrow 5x+20=180 \Rightarrow 5x=160 \Rightarrow x=32.
Answer: 32. (Option 2)

140) If 5^{x-1}+5^x+5^{x+1}=775, then the value of ‘x’ is……

A) 1
B) 2
C) 3
D) 4

View Answer
D) 4

141) Top view of the figure made up of Cubes joined together is……
NMMS 2023 TS

A) (1)
B) (2)
C) (3)
D) (4)

View Answer
D) (4)

142) The value of \left(1^3+2^3+3^3+4^3\right)^\frac{-3}2 is…….

A) \frac1{100}
B) 100
C) \frac1{1000}
D) 1000

View Answer
B) 100

143) ABCD and DBEF are rectangles drawn as shown in the figure. If AB 4cm and AD = 3cm, then area of rectangle DBEF (in square cm) is…….
NMMS 2023 TS

A) 10
B) 12
C) 14
D) 15

View Answer
D) 15

144) Marked price of an article is ₹ 675. If it is sold at a discount of 20%, then it’s selling price (in rupees) is….

A) 540
B) 525
C) 510
D) 500

View Answer
A) 540

Explanation:Marked price = ₹675, discount = 20\%. Selling price = 0.8\times675.
0.8\times675 = 540.
Answer: ₹540. (Option 1)

145) The compound interest on ₹ 8000 at 5% per annum for 2 years compounded annually is……

A) ₹ 8280
B) ₹ 400
C) ₹ 8420
D) ₹ 820

View Answer
D) ₹ 820

Explanation:Compound interest on ₹8000 at 5\% p.a. for 2 years (compounded annually).
Amount =8000(1+0.05)^2 =8000(1.05)^2 =8000(1.1025)=8820.
Compound interest =8820-8000=820.
Answer: ₹820. (Option 4)

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