SSC CPO Paper I · 6,600 questions

SSC CPO Paper I 2020, 2022, 2023, 2024 PYQs

Real previous year questions, subject-wise. Try the question, then open the answer and the explanation in Hindi & English – free, no login.

Complete solved eBook Mock tests

6,600 questions.

Q5541.

A ladder is resting against a wall. The angle between the foot of the ladder and the wall is 45° and the foot of the ladder is 6.6 m away from the wall. The length of the ladder (in m) is:

एक सीढ़ी दीवार के सहारे टिकी हुई है। सीढ़ी के पाद और दीवार के बीच का कोण 45° है और सीढ़ी का पाद दीवार से 6.6 m दूर है। सीढ़ी की लंबाई (m में) है:

A $6.6 \times \sqrt{2}$
B $3.3 \times \sqrt{2}$
C $2.2 \times \sqrt{2}$
D $3.6 \times \sqrt{2}$

Show answer

Concept: Ladder, wall and ground form a right triangle; at 45° it is isosceles.

Angle with the wall $= 45^{\circ}$ , so angle with the ground is also $45^{\circ}$ .
$\cos 45^{\circ} = \frac{6.6}{L} \Rightarrow L = 6.6 \times \sqrt{2}$ m.

Trick: 45-45-90 triangle: hypotenuse = leg $\times \sqrt{2}$ .

Answer: (A) $6.6 \times \sqrt{2}$

अवधारणा: सीढ़ी, दीवार और ज़मीन समकोण त्रिभुज बनाते हैं; 45° पर यह समद्विबाहु है।

दीवार से कोण $45^{\circ}$ , अतः ज़मीन से भी $45^{\circ}$ ।
$\cos 45^{\circ} = \frac{6.6}{L} \Rightarrow L = 6.6 \times \sqrt{2}$ m।

ट्रिक: 45-45-90 त्रिभुज: कर्ण = भुजा $\times \sqrt{2}$ ।

उत्तर: (A) $6.6 \times \sqrt{2}$

SSC CPO Paper IQuantitative AptitudeHeights and DistancesSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5542.

The HCF of two numbers is 29, and the other two factors of their LCM are 15 and 13. The smaller of the two numbers is:

दो संख्याओं का महत्तम समापवर्तक (HCF) 29 है, और उनके लघुत्तम समापवर्त्य (LCM) के अन्य दो गुणनखंड 15 और 13 हैं। दोनों संख्याओं में से छोटी संख्या है:

A377
B406
C435
D464

Show answer

Concept: Numbers = HCF × each co-prime factor of the LCM.

Numbers $= 29 \times 13 = 377$ and $29 \times 15 = 435$ .
Smaller number = 377.

Trick: Multiply HCF by the smaller other factor.

Answer: (A) 377

अवधारणा: संख्याएँ = HCF × LCM के सह-अभाज्य गुणनखंड।

संख्याएँ $= 29 \times 13 = 377$ और $29 \times 15 = 435$ ।
छोटी संख्या = 377।

ट्रिक: HCF को छोटे गुणनखंड से गुणा करें।

उत्तर: (A) 377

SSC CPO Paper IQuantitative AptitudeHCF and LCMSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5543.

Solve for

θ

\cos^{2} θ - \sin^{2} θ = \frac{1}{2}

0^{\circ} < θ < 90^{\circ}

θ

के लिए हल कीजिए:

\cos^{2} θ - \sin^{2} θ = \frac{1}{2}

0^{\circ} < θ < 90^{\circ}

।

A $45^{\circ}$
B $60^{\circ}$
C $30^{\circ}$
D $40^{\circ}$

Show answer

Concept: $\cos^{2} θ - \sin^{2} θ = \cos 2 θ$ .

$\cos 2 θ = \frac{1}{2} \Rightarrow 2 θ = 60^{\circ}$ .
$θ = 30^{\circ}$ .

Trick: Check: $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ ✓ at $30^{\circ}$ .

Answer: (C) $30^{\circ}$

अवधारणा: $\cos^{2} θ - \sin^{2} θ = \cos 2 θ$ ।

$\cos 2 θ = \frac{1}{2} \Rightarrow 2 θ = 60^{\circ}$ ।
$θ = 30^{\circ}$ ।

ट्रिक: जाँच: $30^{\circ}$ पर $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ ✓।

उत्तर: (C) $30^{\circ}$

SSC CPO Paper IQuantitative AptitudeTrigonometrySSC CPO Paper I 2020 · 25 Nov Shift 3

Q5544.

On the marked price of ₹1,250 of an article, three successive discounts of 5%, 15% and 20% were offered. The amount (in ₹) of discount received by a customer is:

किसी वस्तु के ₹1,250 के अंकित मूल्य पर 5%, 15% और 20% की तीन क्रमिक छूटें दी गईं। एक ग्राहक को प्राप्त छूट की राशि (₹ में) है:

A442.50
B950.25
C450
D807.50

Show answer

Concept: Apply each discount on the reduced price.

SP $= 1250 \times 0.95 \times 0.85 \times 0.80 = 807.50$ .
Discount $= 1250 - 807.50 = 442.50$ .

Trick: 807.50 is the selling price — a trap option; the question asks for the discount.

Answer: (A) ₹442.50

अवधारणा: हर छूट घटे हुए मूल्य पर लगती है।

SP $= 1250 \times 0.95 \times 0.85 \times 0.80 = 807.50$ ।
छूट $= 1250 - 807.50 = 442.50$ ।

ट्रिक: 807.50 विक्रय मूल्य है — प्रश्न छूट पूछ रहा है।

उत्तर: (A) ₹442.50

SSC CPO Paper IQuantitative AptitudeSuccessive DiscountsSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5545.

\cot θ = \frac{1}{\sqrt{3}}

0^{\circ} < θ < 90^{\circ}

, then the value of

\frac{2 - \sin^{2} θ}{1 - \cos^{2} θ} + ({cosec}^{2} θ - \sec θ)

is:

यदि

\cot θ = \frac{1}{\sqrt{3}}

0^{\circ} < θ < 90^{\circ}

है, तो

\frac{2 - \sin^{2} θ}{1 - \cos^{2} θ} + ({cosec}^{2} θ - \sec θ)

का मान है:

Show answer

Concept: $\cot θ = \frac{1}{\sqrt{3}} \Rightarrow θ = 60^{\circ}$ .

$\sin^{2} θ = \frac{3}{4}$ , $\cos^{2} θ = \frac{1}{4}$ : $\frac{2 - \frac{3}{4}}{1 - \frac{1}{4}} = \frac{5}{3}$ .
${cosec}^{2} θ = \frac{4}{3}$ , $\sec θ = 2$ .
Value $= \frac{5}{3} + \frac{4}{3} - 2 = 3 - 2 = 1$ .

Trick: Note $1 - \cos^{2} θ = \sin^{2} θ = \frac{3}{4}$ .

Answer: (D) 1

अवधारणा: $\cot θ = \frac{1}{\sqrt{3}} \Rightarrow θ = 60^{\circ}$ ।

$\frac{2 - \frac{3}{4}}{\frac{3}{4}} = \frac{5}{3}$ ।
${cosec}^{2} θ = \frac{4}{3}$ , $\sec θ = 2$ ।
मान $= \frac{5}{3} + \frac{4}{3} - 2 = 1$ ।

ट्रिक: $1 - \cos^{2} θ = \sin^{2} θ$ ।

उत्तर: (D) 1

SSC CPO Paper IQuantitative AptitudeTrigonometrySSC CPO Paper I 2020 · 25 Nov Shift 3

Q5546.

Two circles of radii 15 cm and 10 cm intersect each other and the length of their common chord is 16 cm. What is the distance (in cm) between their centres?

15 cm और 10 cm त्रिज्या वाले दो वृत्त एक-दूसरे को प्रतिच्छेद करते हैं और उनकी उभयनिष्ठ जीवा की लंबाई 16 cm है। उनके केंद्रों के बीच की दूरी (cm में) कितनी है?

A $12 + 3 \sqrt{7}$
B $15 + 2 \sqrt{161}$
C $10 + \sqrt{161}$
D $6 + \sqrt{161}$

Show answer

Concept: The line of centres is the perpendicular bisector of the common chord.

Half chord AM $= 8$ cm.
$O_{1} M = \sqrt{15^{2} - 8^{2}} = \sqrt{161}$ ; $O_{2} M = \sqrt{10^{2} - 8^{2}} = 6$ .
$O_{1} O_{2} = 6 + \sqrt{161}$ cm.

Trick: Use Pythagoras in each circle separately with the half chord.

Answer: (D) $6 + \sqrt{161}$

अवधारणा: केंद्रों को मिलाने वाली रेखा उभयनिष्ठ जीवा (common chord) का लंब समद्विभाजक है।

आधी जीवा AM $= 8$ cm।
$O_{1} M = \sqrt{225 - 64} = \sqrt{161}$ ; $O_{2} M = \sqrt{100 - 64} = 6$ ।
$O_{1} O_{2} = 6 + \sqrt{161}$ cm।

ट्रिक: हर वृत्त में आधी जीवा से पाइथागोरस लगाएँ।

उत्तर: (D) $6 + \sqrt{161}$

SSC CPO Paper IQuantitative AptitudeGeometry – Intersecting CirclesSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5547.

The fasting blood sugar level (F) and the post prandial blood sugar level (PP) of a patient was monitored for four weeks W1, W2, W3, W4, and the readings (in mg/dl) for the four weeks are as follows:
F/PP: 170/220, 150/180, 120/150, 90/120.
These have been presented through a bar graph. The normal range is considered to be:
F/PP: 70 – 100/100 – 130.
(Bar graph data)

Week	F	PP
W1	170	220
W2	150	180
W3	120	150
W4	90	120

During which week was the fasting blood sugar level of the patient approximately 115.4% above the upper limit of normal PP blood sugar level?

चार सप्ताहों W1, W2, W3, W4 तक एक रोगी के उपवास (fasting) रक्त शर्करा स्तर (F) और भोजन के पश्चात (post prandial) रक्त शर्करा स्तर (PP) की निगरानी की गई, और चार सप्ताहों की रीडिंग (mg/dl में) इस प्रकार हैं:
F/PP: 170/220, 150/180, 120/150, 90/120.
इन्हें एक दंड आलेख के माध्यम से प्रस्तुत किया गया है। सामान्य सीमा इस प्रकार मानी जाती है:
F/PP: 70 – 100/100 – 130.
(दंड आलेख के आँकड़े)

सप्ताह	F	PP
W1	170	220
W2	150	180
W3	120	150
W4	90	120

किस सप्ताह के दौरान रोगी का उपवास रक्त शर्करा स्तर, सामान्य PP रक्त शर्करा स्तर की ऊपरी सीमा से लगभग 115.4% अधिक था?

Show answer

Concept: Upper limit of normal PP = 130; compare each week's F value with 130 as a percentage.

W2: $\frac{150}{130} \times 100 \approx 115.4 %$ — F equals 115.4% of 130 (15.4% higher).
Others: W1 $\frac{170}{130} \approx 130.8 %$ , W3 $\frac{120}{130} \approx 92.3 %$ , W4 $\frac{90}{130} \approx 69.2 %$ .
Only W2 gives the 115.4% figure.

Trick: $\frac{15}{13} = 1.1538$ — recognise the fraction behind 115.4%.

Answer: (C) W2

अवधारणा: सामान्य PP की ऊपरी सीमा = 130; हर सप्ताह के F मान को 130 के प्रतिशत में देखें।

W2: $\frac{150}{130} \times 100 \approx 115.4 %$ — F, 130 का 115.4% है (15.4% अधिक)।
W1 $\approx 130.8 %$ , W3 $\approx 92.3 %$ , W4 $\approx 69.2 %$ ।
केवल W2 से 115.4% आता है।

ट्रिक: $\frac{15}{13} = 1.1538$ — 115.4% के पीछे की भिन्न पहचानें।

उत्तर: (C) W2

SSC CPO Paper IQuantitative AptitudeBar Graph (DI)SSC CPO Paper I 2020 · 25 Nov Shift 3

Q5548.

Study the given histogram that shows the marks obtained by students in an examination and answer the question that follows.
(Histogram data: marks on the horizontal axis, number of students on the vertical axis)

Marks	Number of Students
150 – 200	30
200 – 250	45
250 – 300	60
300 – 350	35
350 – 400	40
400 – 450	35

The number of students who obtained less than 200 marks is:

दिए गए आयतचित्र (histogram) का अध्ययन कीजिए, जो किसी परीक्षा में छात्रों द्वारा प्राप्त अंकों को दर्शाता है, और उसके बाद दिए गए प्रश्न का उत्तर दीजिए।
(आयतचित्र के आँकड़े: क्षैतिज अक्ष पर अंक, ऊर्ध्वाधर अक्ष पर छात्रों की संख्या)

अंक	छात्रों की संख्या
150 – 200	30
200 – 250	45
250 – 300	60
300 – 350	35
350 – 400	40
400 – 450	35

200 से कम अंक प्राप्त करने वाले छात्रों की संख्या है:

A75
B30
C45
D135

Show answer

Concept: Less than 200 marks means only the class 150–200.

Class 150–200 has 30 students.
No class starts below 150, so the answer is 30.

Trick: 75 (= 30 + 45) is the count below 250, a trap.

Answer: (B) 30

अवधारणा: 200 से कम अंक = केवल 150–200 वाली class।

150–200 में 30 छात्र।
150 से नीचे कोई class नहीं, अतः उत्तर 30।

ट्रिक: 75 (= 30 + 45) 250 से कम वालों की संख्या है — जाल विकल्प।

उत्तर: (B) 30

SSC CPO Paper IQuantitative AptitudeHistogram (DI)SSC CPO Paper I 2020 · 25 Nov Shift 3

Q5549.

The value of

7 \div [5 + 1 \div 2 - {4 + (4 of 2 \div 4) + (5 \div 5 of 2)}]

is:

7 \div [5 + 1 \div 2 - {4 + (4 of 2 \div 4) + (5 \div 5 of 2)}]

का मान है:

A7
B $\frac{7}{2}$
C−7
D $- \frac{7}{2}$

Show answer

Concept: Solve innermost brackets first; 'of' before ÷.

$4 of 2 \div 4 = 8 \div 4 = 2$ ; $5 \div 5 of 2 = 5 \div 10 = \frac{1}{2}$ .
Curly bracket: $4 + 2 + \frac{1}{2} = \frac{13}{2}$ .
Square bracket: $5 + \frac{1}{2} - \frac{13}{2} = - 1$ .
$7 \div (- 1) = - 7$ .

Trick: '5 ÷ 5 of 2' is $5 \div 10$ , not $1 \times 2$ .

Answer: (C) $- 7$

अवधारणा: पहले अंदर के कोष्ठक; 'of' को ÷ से पहले।

$4 of 2 \div 4 = 2$ ; $5 \div 5 of 2 = 5 \div 10 = \frac{1}{2}$ ।
मझला कोष्ठक: $\frac{13}{2}$ ।
बड़ा कोष्ठक: $5 + \frac{1}{2} - \frac{13}{2} = - 1$ ।
$7 \div (- 1) = - 7$ ।

ट्रिक: '5 ÷ 5 of 2' $= 5 \div 10$ होता है।

उत्तर: (C) $- 7$

SSC CPO Paper IQuantitative AptitudeSimplification – BODMASSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5550.

Two pipes A and B can fill a cistern in

12 \frac{1}{2}

hours and 25 hours, respectively. The pipes are opened simultaneously and it is found that due to a leakage in the bottom, it took 1 hour 40 minutes more to fill the cistern. When the cistern is full, in how much time will the leak empty the cistern?

दो पाइप A और B किसी टंकी को क्रमशः

12 \frac{1}{2}

घंटे और 25 घंटे में भर सकते हैं। पाइपों को एक साथ खोला जाता है और यह पाया जाता है कि तल में एक रिसाव के कारण टंकी को भरने में 1 घंटा 40 मिनट अधिक लगे। टंकी के पूरी भरी होने पर, रिसाव उसे कितने समय में खाली कर देगा?

A45 hours45 घंटे
B42 hours42 घंटे
C48 hours48 घंटे
D50 hours50 घंटे

Show answer

Concept: Leak rate = combined filling rate − actual net rate.

A + B per hour $= \frac{2}{25} + \frac{1}{25} = \frac{3}{25}$ ⇒ time $= \frac{25}{3}$ h = 8 h 20 min.
With leak: 8 h 20 min + 1 h 40 min = 10 h ⇒ net rate $= \frac{1}{10}$ .
Leak rate $= \frac{3}{25} - \frac{1}{10} = \frac{1}{50}$ ⇒ 50 hours.

Trick: Leak time $= \frac{t_{1} t_{2}}{t_{2} - t_{1}} = \frac{\frac{25}{3} \times 10}{\frac{5}{3}} = 50$ .

Answer: (D) 50 hours

अवधारणा: रिसाव दर = संयुक्त भरने की दर − वास्तविक शुद्ध दर।

A + B प्रति घंटा $= \frac{3}{25}$ ⇒ समय $= \frac{25}{3}$ h = 8 h 20 min।
रिसाव के साथ 10 h ⇒ शुद्ध दर $= \frac{1}{10}$ ।
रिसाव दर $= \frac{3}{25} - \frac{1}{10} = \frac{1}{50}$ ⇒ 50 घंटे।

ट्रिक: रिसाव समय $= \frac{t_{1} t_{2}}{t_{2} - t_{1}} = 50$ ।

उत्तर: (D) 50 घंटे

SSC CPO Paper IQuantitative AptitudePipes and CisternsSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5551.

Select the most appropriate option to substitute the underlined segment in the given sentence. If there is no need to substitute it, select ‘No substitution required’.
The exam was not so easy that we had expected.

Aas easy as
BNo substitution required
Cas easy so
Dsuch easy as

Show answer

Concept: Equal comparison uses as + adjective + as (negative may also use so … as). So … that shows result, not comparison.

The sentence compares the exam with what we had expected → a comparison structure is needed.
so easy that needs a result clause, not 'we had expected' → wrong.
as easy so and such easy as are ungrammatical pairs.
Correct: The exam was not as easy as we had expected.

Trick: Comparison of equality = as … as; result = so … that.

Answer: (A) as easy as

अवधारणा: समानता की तुलना में as + adjective + as लगता है। so … that परिणाम (result) बताता है, तुलना नहीं।

वाक्य में exam की तुलना हमारी अपेक्षा से की गई है → तुलना की संरचना चाहिए।
so easy that के बाद परिणाम वाला clause आना चाहिए → गलत।
as easy so तथा such easy as गलत जोड़े हैं।
सही: The exam was not as easy as we had expected.

ट्रिक: तुलना = as … as; परिणाम = so … that।

उत्तर: (A) as easy as

SSC CPO Paper IEnglish ComprehensionSentence Improvement – ComparisonSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5552.

Select the most appropriate meaning of the given idiom.
High time

Awell in time for something
Btime to ﬂy high
Cpast the appropriate time
Dtime to celebrate something

Show answer

Concept: High time means the right time for something has already passed; it should have been done earlier (e.g. It is high time we left).

The idiom expresses urgency/lateness, not punctuality → 'well in time' is the opposite.
It has nothing to do with flying or celebrating.
So it means 'past the appropriate time'.

Yaad rakhein: 'It's high time + past tense' → already late, act now.

Answer: (C) past the appropriate time

अवधारणा: High time का अर्थ है – किसी काम का उचित समय निकल चुका है, उसे पहले ही हो जाना चाहिए था (जैसे It is high time we left)।

यह मुहावरा देर/तात्कालिकता बताता है; 'well in time' इसका उल्टा है।
उड़ने या उत्सव से इसका कोई संबंध नहीं।
अतः अर्थ = 'past the appropriate time' (उचित समय बीत जाना)।

याद रखें: 'It's high time + past tense' → पहले ही देर हो चुकी है।

उत्तर: (C) past the appropriate time

SSC CPO Paper IEnglish ComprehensionIdiomsSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5553.

Identify the segment in the sentence which contains a grammatical error.
He is over eighty and dare not read without glasses.

Aover eighty and
Bdare not read
CHe is
Dwithout glasses

Show answer

Concept: Dare not means 'not brave enough to'. To show lack of ability (because of weak eyesight) we use cannot.

A man over eighty is unable to read without glasses – it is a matter of ability, not courage.
So dare not read gives the wrong sense; it should be cannot read.
The other segments are correct.

Trick: Courage → dare; ability → can/cannot.

Answer: (B) dare not read

अवधारणा: Dare not का अर्थ है 'हिम्मत न होना'। क्षमता न होने (कमज़ोर नज़र) के लिए cannot का प्रयोग होता है।

अस्सी वर्ष से अधिक का व्यक्ति चश्मे के बिना पढ़ नहीं सकता – यह क्षमता की बात है, साहस की नहीं।
इसलिए dare not read गलत अर्थ देता है; सही है cannot read।
बाकी भाग सही हैं।

ट्रिक: साहस → dare; क्षमता → can/cannot।

उत्तर: (B) dare not read

SSC CPO Paper IEnglish ComprehensionError Spotting – ModalsSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5554.

Select the most appropriate option to substitute the underlined segment in the given sentence. If there is no need to substitute it, select ‘No substitution required’.
Suman performed at the festival well yesterday.

Ayesterday at the festival well
Bwell at the festival yesterday
CNo substitution required
Dat the festival yesterday well

Show answer

Concept: Adverbs after the verb follow the order Manner → Place → Time (MPT).

Manner: well; Place: at the festival; Time: yesterday.
Correct order: performed well at the festival yesterday.
All other options break the MPT order.

Trick: MPT – 'How, Where, When'.

Answer: (B) well at the festival yesterday

अवधारणा: क्रिया के बाद adverbs का क्रम होता है Manner → Place → Time (MPT)।

Manner: well; Place: at the festival; Time: yesterday।
सही क्रम: performed well at the festival yesterday।
बाकी विकल्प MPT क्रम तोड़ते हैं।

ट्रिक: MPT – कैसे, कहाँ, कब।

उत्तर: (B) well at the festival yesterday

SSC CPO Paper IEnglish ComprehensionSentence Improvement – Word OrderSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5555.

Select the most appropriate synonym of the given word.
OBLIVIOUS

Asensitive
Bobvious
Cignorant
Dattentive

Show answer

Concept: Oblivious = not aware of or not concerned about what is happening around.

ignorant = lacking knowledge/awareness → same sense.
sensitive, attentive are nearly opposites.
obvious (clear) only looks similar in spelling.

Trick: Oblivious ≈ unaware ≈ ignorant.

Answer: (C) ignorant

अवधारणा: Oblivious = आस-पास क्या हो रहा है उससे अनजान/बेख़बर।

ignorant = अनजान → समान अर्थ।
sensitive, attentive लगभग विपरीत हैं।
obvious (स्पष्ट) केवल वर्तनी में मिलता-जुलता है।

ट्रिक: Oblivious ≈ unaware ≈ ignorant (बेख़बर)।

उत्तर: (C) ignorant

SSC CPO Paper IEnglish ComprehensionSynonymsSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5556.

Select the most appropriate meaning of the given idiom.
Apple of discord

Asomething very special
Bsomething that causes friction
Ca good opportunity
Da tempting offer

Show answer

Concept: Apple of discord = the root cause of a quarrel or dispute (from the Greek myth of the golden apple that led to the Trojan War).

'Discord' means disagreement/conflict.
So the idiom means something that causes friction.
'Something very special' is the meaning of 'apple of one's eye' – a trap.

Trick: Apple of discord = cause of fight; apple of one's eye = dearest.

Answer: (B) something that causes friction

अवधारणा: Apple of discord = झगड़े की जड़ (यूनानी कथा के सुनहरे सेब से, जिससे ट्रॉय युद्ध हुआ)।

'Discord' = मतभेद/कलह।
अतः अर्थ = जो टकराव (friction) पैदा करे।
'Something very special' तो 'apple of one's eye' का अर्थ है – जाल विकल्प।

ट्रिक: Apple of discord = झगड़े की जड़; apple of one's eye = सबसे प्रिय।

उत्तर: (B) something that causes friction

SSC CPO Paper IEnglish ComprehensionIdiomsSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5557.

Select the most appropriate one-word substitution for the given group of words.
To cross streets on foot carelessly

Acatwalk
Bjaywalk
Cwander
Dstroll

Show answer

Concept: Jaywalk = to cross or walk in a street carelessly, ignoring traffic rules.

catwalk = narrow platform for fashion shows.
wander, stroll = walk aimlessly/leisurely, not about crossing streets.
So the answer is jaywalk.

Yaad rakhein: Jaywalker = careless road-crosser.

Answer: (B) jaywalk

अवधारणा: Jaywalk = यातायात नियमों की अनदेखी कर लापरवाही से सड़क पार करना।

catwalk = फैशन शो का रैंप।
wander, stroll = घूमना/टहलना, सड़क पार करने से संबंध नहीं।
अतः उत्तर jaywalk।

याद रखें: Jaywalker = लापरवाही से सड़क पार करने वाला।

उत्तर: (B) jaywalk

SSC CPO Paper IEnglish ComprehensionOne Word SubstitutionSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5558.

Identify the segment in the sentence which contains a grammatical error.
I am suitably dressed for the occasion, am I?

Aam I
BI am
Csuitably dressed
Dfor the occasion

Show answer

Concept: A positive statement takes a negative tag. For 'I am', the negative tag is aren't I?

Statement 'I am suitably dressed…' is positive.
So the tag must be negative: aren't I?, not am I?
Error lies in segment 'am I'.

Trick: Positive → negative tag; I am → aren't I.

Answer: (A) am I

अवधारणा: सकारात्मक वाक्य के साथ नकारात्मक tag आता है। 'I am' का negative tag aren't I? होता है।

वाक्य 'I am suitably dressed…' सकारात्मक है।
इसलिए tag नकारात्मक होगा: aren't I?, न कि am I?
त्रुटि 'am I' में है।

ट्रिक: Positive → negative tag; I am → aren't I।

उत्तर: (A) am I

SSC CPO Paper IEnglish ComprehensionError Spotting – Question TagsSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5559.

Select the most appropriate one-word substitution for the given group of words.
Downfall that satisﬁes natural justice

Aparadox
Bclimax
Ccatharsis
Dnemesis

Show answer

Concept: Nemesis = a downfall or punishment that is deserved and cannot be escaped (natural justice); named after the Greek goddess of retribution.

paradox = self-contradictory statement.
climax = highest point; catharsis = release of emotions.
Only nemesis fits 'deserved downfall'.

Yaad rakhein: Nemesis – goddess of revenge → deserved ruin.

Answer: (D) nemesis

अवधारणा: Nemesis = ऐसा पतन/दंड जो उचित हो और जिससे बचा न जा सके (प्राकृतिक न्याय); यूनानी प्रतिशोध की देवी के नाम पर।

paradox = विरोधाभास।
climax = चरम बिंदु; catharsis = भावनाओं का निकास।
केवल nemesis 'उचित पतन' बताता है।

याद रखें: Nemesis – प्रतिशोध की देवी → उचित विनाश।

उत्तर: (D) nemesis

SSC CPO Paper IEnglish ComprehensionOne Word SubstitutionSSC CPO Paper I 2020 · 25 Nov Shift 3

Q5560.

Select the most appropriate meaning of the given idiom.
To take stock of

Ato buy stocks
Bto assess a situation
Cto check the accounts
Dto store goods

Show answer

Concept: To take stock of = to review or assess a situation carefully before deciding what to do.

Though 'stock' suggests goods or shares, the idiom is figurative.
e.g. After the flood, officials took stock of the damage.
So it means 'to assess a situation'.

Trick: Take stock = take a careful look.

Answer: (B) to assess a situation

अवधारणा: To take stock of = कोई निर्णय लेने से पहले स्थिति का सावधानी से आकलन करना।

'Stock' से माल या शेयर लगते हैं, पर मुहावरा लाक्षणिक है।
जैसे After the flood, officials took stock of the damage.
अतः अर्थ = स्थिति का आकलन करना।

ट्रिक: Take stock = ध्यान से जायज़ा लेना।

उत्तर: (B) to assess a situation

SSC CPO Paper IEnglish ComprehensionIdiomsSSC CPO Paper I 2020 · 25 Nov Shift 3